InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your Class 11 knowledge and support exam preparation. Choose a topic below to get started.
| 49301. |
The equilibrium constant of a reaction is 73. Calculate standard free energy change. |
| Answer» <html><body><p>Both A & R are true and R is the correct <a href="https://interviewquestions.tuteehub.com/tag/explanation-455162" style="font-weight:bold;" target="_blank" title="Click to know more about EXPLANATION">EXPLANATION</a> of A <br/>Both A & R are true but R is not the correct explanation of A <br/>A is true but R is <a href="https://interviewquestions.tuteehub.com/tag/false-459184" style="font-weight:bold;" target="_blank" title="Click to know more about FALSE">FALSE</a> <br/>A is false but R is true </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :D</body></html> | |
| 49302. |
(a) Stability of a crystal is reflected in the magnitude of its melting point'. Comment (b)The melting point of some compounds are given below: Water = 273 K, Ethyl alcohol =155.7 K, Diethyl ether =156.8 K, Methane =90.5 K. What can you say about the intermolecular forces between these molecules ? |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :(a) Higher the <a href="https://interviewquestions.tuteehub.com/tag/melting-1093090" style="font-weight:bold;" target="_blank" title="Click to know more about MELTING">MELTING</a> point, greater are the forces holding the constituent particles together and hence greater is the stability<br/> (b) The intermolecular forces in water and ethyl alcohol are mainly the hydrogen bonding Higher melting point of water than alcohol shows that hydrogen bonding in ethyl alcohol molecules is not a strong as in water molecules. <a href="https://interviewquestions.tuteehub.com/tag/diethyl-951375" style="font-weight:bold;" target="_blank" title="Click to know more about DIETHYL">DIETHYL</a> ether is a polar molecule. The intermolecular forces present in them are dipole dipole attraction Methane is a non-polar molecule. The only forces present in them are the weak van <a href="https://interviewquestions.tuteehub.com/tag/der-948864" style="font-weight:bold;" target="_blank" title="Click to know more about DER">DER</a> Waals forces (London dispersion forces)</body></html> | |
| 49303. |
(a) ' Stability of a crystal is reflected in the magnitude of its melting point . Comment (b) The melting points of some compounds are given below : water = 273 K ,Ethyl alcohal= 155.7 K , Diethyl ether = 156.8 K , Methane = 90.5kK what can you say about the intermolecular forces between these molescular ? |
| Answer» <html><body><p></p>Solution :(a) Higher the <a href="https://interviewquestions.tuteehub.com/tag/meting-2830889" style="font-weight:bold;" target="_blank" title="Click to know more about METING">METING</a> point, greater are the forces holding the consituent particles together and hence greater is the stability. <br/> (b)The intermolecular forces in water and entyl alcohalare mainly the hydrogen <a href="https://interviewquestions.tuteehub.com/tag/bonding-900657" style="font-weight:bold;" target="_blank" title="Click to know more about BONDING">BONDING</a>. Higher <a href="https://interviewquestions.tuteehub.com/tag/melting-1093090" style="font-weight:bold;" target="_blank" title="Click to know more about MELTING">MELTING</a> point of water than alcoholshows that hydrogen bonding in <a href="https://interviewquestions.tuteehub.com/tag/ethyl-976035" style="font-weight:bold;" target="_blank" title="Click to know more about ETHYL">ETHYL</a> alcohol molecules is not are strong as in water molecules. Diethyl ether is a <a href="https://interviewquestions.tuteehub.com/tag/poler-7708130" style="font-weight:bold;" target="_blank" title="Click to know more about POLER">POLER</a> molecule. The intermolecular forces present in them are dipolediple attraction. Methane is a non-polar molecular.The only forces present in them are the weak van der Waal's forces ( London dispersion forces).</body></html> | |
| 49304. |
A square planar complex is formed by hybridisation of which of the following atomic orbitals ? |
| Answer» <html><body><p><<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>>`s,p_x,p_y ,d_(<a href="https://interviewquestions.tuteehub.com/tag/yz-748889" style="font-weight:bold;" target="_blank" title="Click to know more about YZ">YZ</a>)`<br/>`s,p_x ,P_y ,d_(x^2 -y^2)`<br/>`s,p,p_y,d_(z^2)`<br/>`s,p_y ,P_z ,d_(<a href="https://interviewquestions.tuteehub.com/tag/xy-747762" style="font-weight:bold;" target="_blank" title="Click to know more about XY">XY</a>)`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 49305. |
A spinal is an important class of oxides consisting of two types of metal ions with oxides ions arranged in CCP layer. The normal spinal has one-eighth of the tetrahedral holes occupied by one type of metal ion and one-half of the octahedral holes occupied by another type of metal ion. such a spinal is formed by Zn^(2+), Al^(3+) and O^(2-) in the tetrahedral holes. if formula of the compound is Zn_(x)Al_(y) O_(z), then find the value of (x + y + z) ? The type of packing generated by type (I) is: |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/hexagonal-485015" style="font-weight:bold;" target="_blank" title="Click to know more about HEXAGONAL">HEXAGONAL</a> close packing<br/>squre close packing<br/>cubic close packing<br/>body centered packing</p>Solution :`O_(4) Zn_((<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)/(<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a>) xx 8) Al_((1)/(2) xx 4)` <br/> So, <a href="https://interviewquestions.tuteehub.com/tag/formula-464310" style="font-weight:bold;" target="_blank" title="Click to know more about FORMULA">FORMULA</a> is `ZnAl_(2)O_(4)`</body></html> | |
| 49306. |
For a spontaneous process, in a reaction |
| Answer» <html><body><p>which <a href="https://interviewquestions.tuteehub.com/tag/needs-1112903" style="font-weight:bold;" target="_blank" title="Click to know more about NEEDS">NEEDS</a> some initaiation <a href="https://interviewquestions.tuteehub.com/tag/like-1073845" style="font-weight:bold;" target="_blank" title="Click to know more about LIKE">LIKE</a> <a href="https://interviewquestions.tuteehub.com/tag/heat-21102" style="font-weight:bold;" target="_blank" title="Click to know more about HEAT">HEAT</a> or energy <br/>which takes place instantaneously <br/>which takes place by itself <br/>takes place by itself or by <a href="https://interviewquestions.tuteehub.com/tag/initiation-20359" style="font-weight:bold;" target="_blank" title="Click to know more about INITIATION">INITIATION</a>. </p>Answer :D</body></html> | |
| 49307. |
A spherical balloon of 21 cm diameter is to be filled with hydrogen at NTP from a cylinder containing the gas at20 atmosphere at 27^(@)C.If the cylinder can hold 2.82 litres of water, calculate the number of balloons that can be filled up. |
| Answer» <html><body><p></p>Solution :Volume of the balloon `=(4)/(3)pi r^(3)=(4)/(3)<a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a>(22)/(7)xx((21)/(2))^(3)=4851 cm^(3)` <br/> Volume of the cylinder =2.82 litres=2820 `cm^(3)` <br/> Pressure =20 <a href="https://interviewquestions.tuteehub.com/tag/atm-364409" style="font-weight:bold;" target="_blank" title="Click to know more about ATM">ATM</a>. Temperature =300 K <br/> Converting this to the volume at NTP, <br/> `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2)):. (20xx2820)/(300)=(1xxV_(2))/(273)"or"V_(2)=51324" cm"^(3)`. <br/> When the pressure in the cylinder is <a href="https://interviewquestions.tuteehub.com/tag/reduced-1181337" style="font-weight:bold;" target="_blank" title="Click to know more about REDUCED">REDUCED</a> to one atmosphere,no more `H_(2)` will be released and hence 2820 `cm^(3)` of `H_(2)` will be left in it. Hence, volume of `H_(2)`used in filling the balloons <br/> `=51324-2820 cm^(3)=<a href="https://interviewquestions.tuteehub.com/tag/4850br4-1880139" style="font-weight:bold;" target="_blank" title="Click to know more about 48504">48504</a> cm^(3)`. <br/> Number of balloons filled =48504/4851=10</body></html> | |
| 49308. |
A spectral line of hydrogen with lambda = 4938 Å belongs to the series |
| Answer» <html><body><p>Lyman<br/>Balmer<br/>Paschen<br/>Pfund</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 49309. |
A specisthat. Donatesthe electronpair pf H^(@)is termedas a / an_______. |
| Answer» <html><body><p> <a href="https://interviewquestions.tuteehub.com/tag/nucleophile-1126013" style="font-weight:bold;" target="_blank" title="Click to know more about NUCLEOPHILE">NUCLEOPHILE</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/base-892693" style="font-weight:bold;" target="_blank" title="Click to know more about BASE">BASE</a> <br/>electrophile<br/>acid</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 49310. |
A specific cloud is formed on antarctia in the winter is called ......... |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :<a href="https://interviewquestions.tuteehub.com/tag/nacreous-7701177" style="font-weight:bold;" target="_blank" title="Click to know more about NACREOUS">NACREOUS</a> <a href="https://interviewquestions.tuteehub.com/tag/clouds-919894" style="font-weight:bold;" target="_blank" title="Click to know more about CLOUDS">CLOUDS</a></body></html> | |
| 49311. |
A species ‘X’ contains 20 protons and 18 electrons. (A) species ‘Y’ contains 18 protons and 18 electrons. What are ‘X’ and ‘Y’ respectively ? |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/ca-375" style="font-weight:bold;" target="_blank" title="Click to know more about CA">CA</a> and Ar<br/>`Ca^(2+)` and `S^(2-)`<br/>`Ca^(2+)` and `CI^(-)`<br/>`Ca^(2+)` and Ar</p>Solution :(i) X has 20 protons `therefore` <a href="https://interviewquestions.tuteehub.com/tag/atomic-2477" style="font-weight:bold;" target="_blank" title="Click to know more about ATOMIC">ATOMIC</a> number = Z = 20`therefore`The atom is Ca But it has <a href="https://interviewquestions.tuteehub.com/tag/18-278913" style="font-weight:bold;" target="_blank" title="Click to know more about 18">18</a> electrons `therefore` X <a href="https://interviewquestions.tuteehub.com/tag/must-2185568" style="font-weight:bold;" target="_blank" title="Click to know more about MUST">MUST</a> be `Ca^(2+)` <br/> (<a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a>) Y has 18 protons `therefore`atomic number = 18 = Z `therefore` The atom is Ar But no. of electron in it are 18 Y must be A</body></html> | |
| 49312. |
A spark plug is not necessary in a diesel engine because |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/diesel-439168" style="font-weight:bold;" target="_blank" title="Click to know more about DIESEL">DIESEL</a> is more <a href="https://interviewquestions.tuteehub.com/tag/volatilebrbr-3264686" style="font-weight:bold;" target="_blank" title="Click to know more about VOLATILE">VOLATILE</a> than petrol<br/>diesel has a lower ignition temperature than petrol<br/>calorific <a href="https://interviewquestions.tuteehub.com/tag/value-1442530" style="font-weight:bold;" target="_blank" title="Click to know more about VALUE">VALUE</a> of diesel is more than that of petrol<br/>None of these</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/due-433472" style="font-weight:bold;" target="_blank" title="Click to know more about DUE">DUE</a> to higher compression ratio in diesel <a href="https://interviewquestions.tuteehub.com/tag/engine-25861" style="font-weight:bold;" target="_blank" title="Click to know more about ENGINE">ENGINE</a> the compressed air gets heated up to such a high temperature that the sprayed fuel catches fire on its own</body></html> | |
| 49313. |
A sparingly soluble salt having general formula A_(x)^(p+) B_(y)^(q-) and molar solubility S is in equilibrium with its saturated solution. Derive a relationship between the solubility and solubility product for such salt. |
| Answer» <html><body><p><<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>></p>Solution :Suppose molar solubility of `A_(<a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a>)^(p+) B_(y)^(<a href="https://interviewquestions.tuteehub.com/tag/q-609558" style="font-weight:bold;" target="_blank" title="Click to know more about Q">Q</a>-)` is S mol `<a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a>^(-1)`. Then <br/> `{:(A_(x)^(p+)B_(y)^(q-) ,hArr,x A^(p+),+ ,yB^(q-),,),(,,x S,,y S,,):}` <br/> `K_(sp) = [ A^(p+)]^(x) [B^(q-)]^(y)=[x S]^(x) [ y S ] ^(y)= x^(x) y^(y) S^(x+y)`</body></html> | |
| 49314. |
A sparingly soluble salt having general formula A_x^(p+) B_y^(q-)and molar solubility S is in equilibrium with its saturated solution. Derive a relationship between solubility and the solubility product for such salt. |
| Answer» <html><body><p></p>Solution :A sparingly soluble salt having general formula `A_x^(p+) B_y^(q-)` . Its <a href="https://interviewquestions.tuteehub.com/tag/molar-562965" style="font-weight:bold;" target="_blank" title="Click to know more about MOLAR">MOLAR</a> solubilityis S mol `L^(-1)` . <br/> Then, `A_x^(p+) B_y^(q-)hArr xA_x^(p+)(<a href="https://interviewquestions.tuteehub.com/tag/aq-883254" style="font-weight:bold;" target="_blank" title="Click to know more about AQ">AQ</a>)+yB_y^(q-)` (aq) <br/> S moles of `A_xB_y` dissolve to give <a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a> moles of `A^(p+)` and y moles of `B^(q-)` <br/> Therefore , solubility <a href="https://interviewquestions.tuteehub.com/tag/product-25523" style="font-weight:bold;" target="_blank" title="Click to know more about PRODUCT">PRODUCT</a> <br/> `(K_(sp))=[A^(p+)]^x [B^(q-)]^y` <br/> `=[xS]^x [yS]^y` <br/> `=x^x y^y S^(x+y)`</body></html> | |
| 49315. |
A sparingly soluble salt gets precipitated only when the product of concentration of its ions in the solution (Q_(sp)) becomes greater than its solubility product. If the solubility of BaSO_(4) in water is 8xx10^(-4)mol dm^(-3) , calculate its solubility in 0.01 mol dm^(-3) of H_(2)SO_(4). |
| Answer» <html><body><p></p>Solution :Solubility of `BaSO_(4) ` in water `(S) = 8 xx 10^(-4) ` <a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a> `dm^(-3)` <a href="https://interviewquestions.tuteehub.com/tag/ltbgt-2804401" style="font-weight:bold;" target="_blank" title="Click to know more about LTBGT">LTBGT</a> `BaSO_(4) hArr Ba^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>) + SO_(4)^(2-)` <br/> `:. K_(sp) ` for `BaSO_(4) = [Ba^(2+)][SO_(4)^(2-)]=(8xx10^(-4))(8xx10^(-4))=64xx10^(-8)` <br/> As `H_(2)SO_(4)` <a href="https://interviewquestions.tuteehub.com/tag/ionizes-7690815" style="font-weight:bold;" target="_blank" title="Click to know more about IONIZES">IONIZES</a> completely as `H_(2)SO_(4) rarr 2H^(+) + SO_(4)^(2-),[SO_(4)^(2-)]` producedfrom 0.01 M `H_(2)SO_(2-)]` produced from 0.01 M `H_(2)SO_(4)=0.01 M `. Thus, if S is the solubility of `BaSO_(4) ` in `H_(2)SO_(4) ` , then <br/> `K_(ap) = [Ba^(2+)][SO_(4)^(2-)] = S (S+ 0.01 ) = 64 xx 10^(-8) `(calculated above ) <br/>or `S^(2)+0.01 S - 64 xx 10^(-8) = 0 ` <br/> `:. S= (-0.01 <a href="https://interviewquestions.tuteehub.com/tag/pm-1156895" style="font-weight:bold;" target="_blank" title="Click to know more about PM">PM</a> sqrt((0.01)^(2)+4xx64xx10^(-8)))/(2) = (-0.01 pm sqrt(10^(-4)+256xx10^(-8)))/(2)` <br/> `=(-0.01 pm sqrt(10^(-4)(1+256xx10^(-2))))/(2) = (-0.01 pm 10^(-2) sqrt(1+256xx10^(-2)))/(2)` <br/> `=(-0.01 pm 10^(-2)sqrt(1.256))/(2) = (-10^(-2)+1.12xx10^(-2))/(2)` <br/> `=((-1+1.12)xx10^(-2))/(2)` <br/> `=((-1+1.12)xx10^(-2))/(2) = (0.12)/(2) xx 10^(-2) = 6 xx 10^(-4) ` mol `dm^(-3)`.</body></html> | |
| 49316. |
A sparingly soluble salt gets precipitated only when the product of concentration of its ions in the solution Q_((sp)) becomes greater than its solubility product. If the solubility of BaSO_4 in water is 8 xx 10^(-4) "mol dm"^(-3). Calculate its solubility in 0.01 mol "dm"^(-3) of H_2SO_4. |
| Answer» <html><body><p></p>Solution :`BaSO_(4(g)) hArr Ba_((aq))^(2+) + SO_(4(aq))^(2-)` <br/> `K_(sp)` for `BaSO_4 = [Ba^(2+)] [ SO_4^(2-) ] =sxxs =S^2` <br/> But `S=<a href="https://interviewquestions.tuteehub.com/tag/8xx10-1931283" style="font-weight:bold;" target="_blank" title="Click to know more about 8XX10">8XX10</a>^(-4) "mol dm"^(-3)` <br/>`<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a> K_(sp)=(8xx10^(-4))^2 = 64xx10^(-8)` <br/> In the presence of 0.01 `MH_2 SO_4`, the expression for `K_(sp)` will be <br/> `K_(sp)=[Ba^(2+)][SO_4^(2-)]`<br/> `K_(sp)`=(s)(s+0.01)(0.01M `SO_4^(2-)` <a href="https://interviewquestions.tuteehub.com/tag/ions-1051295" style="font-weight:bold;" target="_blank" title="Click to know more about IONS">IONS</a> from 0.01 M `H_2SO_4` ) <br/>64 x 104 = s. (s+0.01) <br/> `s^2=0.01s-64xx10^(-8)=0` <br/> `s=(-0.01pmsqrt((0.01)^2+(4xx64xx10^(-8))))/2` <br/> `=(-0.01pm sqrt(10^(-4)+256xx10^(-8)))/2` <br/> `=(-0.01 <a href="https://interviewquestions.tuteehub.com/tag/pm-1156895" style="font-weight:bold;" target="_blank" title="Click to know more about PM">PM</a> sqrt(10^(-4)(1+256 xx10^(-4))))/2` <br/> `=(-0.01 pm 10^(-2) sqrt(1+0.0256))/2` <br/> `=(10^(-2) (-1pm1.012719))/2` <br/> `=5xx10^(-3) (-1+1.012719)` <br/> `=6.4xx10^(-5) "mol dm"^(-3)`</body></html> | |
| 49317. |
Asolution with pH 2.699 is diluted two times, then calculate pH of the resulting solution. [Given antilog of 0.3010 = 2 |
| Answer» <html><body><p><br/></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/new-1114486" style="font-weight:bold;" target="_blank" title="Click to know more about NEW">NEW</a> pH = 2.699 + `log (<a href="https://interviewquestions.tuteehub.com/tag/2v-300496" style="font-weight:bold;" target="_blank" title="Click to know more about 2V">2V</a>)/( <a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a>) =<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>`</body></html> | |
| 49318. |
A solution which remains in equilibrium with undissolved solute is said to be saturated. The concentration of a saturated solution at a given temperature is called solubility. The product of concentration of ions in a saturated solution of an electrolyte at a given temperature, is called solubility product (K_(sp)). For the electrolyte, A_(x),B_(y),:A_(x),B_(y(s)) rarr xA^(y+)+ y^(Bx-), with solubility S, the solubility product (K_(sp)) =x^(x)xxy^(y) xx s^(x+y). While calculating the solubility of a sparingly soluble salt in the presence of some strong electrolyte containing a common ion, the common ion concentration is practically equal to that of strong electrolyte. If in a solution, the ionic product of an clectrolyte exceeds its K_(sp), value at a particular temperature, then precipitation occurs. The solubility of BaSO_(4), in 0.1 M BaCl_(2), solution is (K_(sp), of BaSO_(4), = 1.5 xx 10^(-9)) |
| Answer» <html><body><p>` <a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>.5 xx 10^(-9) M` <br/>`1.5 xx10^(-8) M` <br/>` 2.25 xx 10^(-<a href="https://interviewquestions.tuteehub.com/tag/16-276476" style="font-weight:bold;" target="_blank" title="Click to know more about 16">16</a>)M` <br/>` 2.25 xx 10^(-18) M` </p>Solution :` BaSO_4 hArr Ba^(+2)+SO_4^(-2) ` <br/> ` <a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>.1 M BaCl_2 =0.1 M Ba^(+2) ` <br/>` [SO_4^(-2)]= S, [Ba^(+2) ]= (0.1 +S) ~~ 0.1M` <br/> ` K_(sp)=[Ba^(+2) ] [SO_4^(-2) ], 1.5 xx 10 ^(-9) =(0.1 ) (S) ` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/rarr-1175461" style="font-weight:bold;" target="_blank" title="Click to know more about RARR">RARR</a> S = 1.5 xx 1 0^(-8) `</body></html> | |
| 49319. |
A solution which remains in equilibrium with undissolved solute is said to be saturated. The concentration of a saturated solution at a given temperature is called solubility. The product of concentration of ions in a saturated solution of an electrolyte at a given temperature, is called solubility product (K_(sp)). For the electrolyte, A_(x),B_(y),:A_(x),B_(y(s)) rarr xA^(y+)+ y^(Bx-), with solubility S, the solubility product (K_(sp)) =x^(x)xxy^(y) xx s^(x+y). While calculating the solubility of a sparingly soluble salt in the presence of some strong electrolyte containing a common ion, the common ion concentration is practically equal to that of strong electrolyte. If in a solution, the ionic product of an clectrolyte exceeds its K_(sp), value at a particular temperature, then precipitation occurs. The solubility of PbSO_(4), in water is 0.303 g/l at 25^(@)C, its solubility product at that temperature is |
| Answer» <html><body><p>` 10^(-4)M^(2) ` <br/>` 9.18 xx 10^(-4) M` <br/>` 10^(-6)M^(2)` <br/>` 9.18 xx 10^(-8) M^(2)` </p>Solution :` S= 0.303 g // lit ` <br/> ` <a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a> . <a href="https://interviewquestions.tuteehub.com/tag/wt-1462289" style="font-weight:bold;" target="_blank" title="Click to know more about WT">WT</a> = 207 + 32 + 64 =303 ` <br/><a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a> ` =( 0.303)/( 303)=10 ^(-3) `<br/>` S = 10 ^(-3)" <a href="https://interviewquestions.tuteehub.com/tag/mole-1100299" style="font-weight:bold;" target="_blank" title="Click to know more about MOLE">MOLE</a> " // "lit"` <br/> ` pbSO_4 hArr Punderset(S) b^(+2)+Sunderset( S) O ^(-2), S =10 ^(-3) ` <br/> ` K_(sp)= ( 10 ^(-3) M)^(2) =10 ^(-6)M^(2) `</body></html> | |
| 49320. |
A solution which is 10^(-3) M each in Mn^(2+),Fe^(2+),Zn^(2+)andHg^(2+) is treated with 10^(-16)M sulphide ion. If K_(sp) od MnS, ZnS and HgS are 10^(-15),10^(-25),10^(-20)and10^(-54) respectively, which one will precipitate first ? |
| Answer» <html><body><p>FeS <br/>MnS<br/>HgS <br/>ZnS</p>Solution :Ionic product in the solution `=<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^(-<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)xx10^(16)=10^(-<a href="https://interviewquestions.tuteehub.com/tag/19-280618" style="font-weight:bold;" target="_blank" title="Click to know more about 19">19</a>)` <br/> The metal sulphide having the lowest solubility will precipitate <a href="https://interviewquestions.tuteehub.com/tag/first-461760" style="font-weight:bold;" target="_blank" title="Click to know more about FIRST">FIRST</a> provided the ionic product is higher than the `K_(<a href="https://interviewquestions.tuteehub.com/tag/sp-1219706" style="font-weight:bold;" target="_blank" title="Click to know more about SP">SP</a>)`. Here, all salts are of the same valence-type. So the sulphide having the lowest `K_(sp)` value will precipitate first provided `K_(sp)lt10^(-19)`, HgS has lowest `K_(sp)` value `(10^(-54))`, so will precipitate first.</body></html> | |
| 49321. |
A solution which is 10^(-3) M each in Mn^(2+), Fe^(2+), Zn^(2+) and Hg^(2+) is treated with 10^(-16) M sulphide ion. If K_(sp) of MnS, FeS, ZnS and HgS are 10^(-15), 10^(-23), 10^(-20) and 10^(-54) respectively, which one will precipitate first ? |
| Answer» <html><body><p>FeS<br/>MgS<br/>HgS<br/>ZnS</p>Solution :`HgS` will be <a href="https://interviewquestions.tuteehub.com/tag/precipitated-2948214" style="font-weight:bold;" target="_blank" title="Click to know more about PRECIPITATED">PRECIPITATED</a> <a href="https://interviewquestions.tuteehub.com/tag/first-461760" style="font-weight:bold;" target="_blank" title="Click to know more about FIRST">FIRST</a> because [its `K_(sp)`] is <a href="https://interviewquestions.tuteehub.com/tag/minimum-561095" style="font-weight:bold;" target="_blank" title="Click to know more about MINIMUM">MINIMUM</a>.</body></html> | |
| 49322. |
A solution of which substance can best be used as both titrant and its own indicator in an oxidation-reduction titration ? |
| Answer» <html><body><p>`I_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)`<br/>NaOCl<br/>`K_(2)Cr_(2)O_(7)`<br/>`KMnO_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :N//A</body></html> | |
| 49323. |
A solution of weak acid was titrated with base NaOH. The equivalence point was reached when 36.12 mL of 0.1M NaOH have been added. Now 18.06mL 0.1M HCI were added to titrated solution, the pH was found to be 4.92. What is K_(a) of acid ? |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :`1.2xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>);`</body></html> | |
| 49324. |
A solution of volume V contains a mole of MCI and b mole of NCI where MOH and NOH are two weak bases having dissociation constants K_(1) and K_(2) respectively. Show that the pH of the solution can be expressed as pH=(1)/(2)log_(10) [(K_(1)K_(2))/(K_(w))xx(V)/(aK_(2)+bK_(1))] |
| Answer» | |
| 49325. |
A solution of specific gravity 1.6 g mL^(-1) is 67% by weight. What will be the % by weight of the solution of same acid if it is diluted to specific gravity 1.2 g mL^(-1)? |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :`29.78%` ;</body></html> | |
| 49326. |
A solution of sodium salt of unknown anion when treated with Magnesium chloride solution gives white precipitate only on boiling. The anion is |
| Answer» <html><body><p>`SO_(4)^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>-)`<br/>`HCO_(3)^(-)`<br/>`CO_(3)^(2-)`<br/>`NO_(3)^(-)`.</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/bicarbonates-896608" style="font-weight:bold;" target="_blank" title="Click to know more about BICARBONATES">BICARBONATES</a> on heating form insoluble carbonates<br/> `MgCl_(2) + 2NaHCO_(3)rarr2NaCl+Mg (HCO_(3))_(2)`<br/>`Mg(HCO_(3))_(2)overset(Delta)rarrMgCO_(3)darr+CO_(2)+H_(2)O`</body></html> | |
| 49327. |
A solution of sodium metal is liquid ammonia is strongly reducing due to the presence of |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/sodium-1215510" style="font-weight:bold;" target="_blank" title="Click to know more about SODIUM">SODIUM</a> atoms<br/>Sodium hydride<br/>Sodium amide<br/><a href="https://interviewquestions.tuteehub.com/tag/solvated-7743184" style="font-weight:bold;" target="_blank" title="Click to know more about SOLVATED">SOLVATED</a> <a href="https://interviewquestions.tuteehub.com/tag/electrons-969138" style="font-weight:bold;" target="_blank" title="Click to know more about ELECTRONS">ELECTRONS</a>.</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :A solution of sodium metal in ammonia <a href="https://interviewquestions.tuteehub.com/tag/contains-11473" style="font-weight:bold;" target="_blank" title="Click to know more about CONTAINS">CONTAINS</a> solvated electrons.</body></html> | |
| 49328. |
A solution of sodium metal in liquid ammonia is strongly reducing due to the presence of.... |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/sodium-1215510" style="font-weight:bold;" target="_blank" title="Click to know more about SODIUM">SODIUM</a> <a href="https://interviewquestions.tuteehub.com/tag/atoms-887421" style="font-weight:bold;" target="_blank" title="Click to know more about ATOMS">ATOMS</a> <br/>sodium hydride<br/> sodium <a href="https://interviewquestions.tuteehub.com/tag/amide-859188" style="font-weight:bold;" target="_blank" title="Click to know more about AMIDE">AMIDE</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/solvated-7743184" style="font-weight:bold;" target="_blank" title="Click to know more about SOLVATED">SOLVATED</a> electrons</p>Answer :B</body></html> | |
| 49329. |
A solution of sodium metal in liquid ammonia is strongly reducing due to the presence of |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/sodium-1215510" style="font-weight:bold;" target="_blank" title="Click to know more about SODIUM">SODIUM</a> atoms<br/>sodium hydride<br/>sodium amide<br/>solvated electrons</p>Answer :D</body></html> | |
| 49330. |
Why does the solution of sodium in liquid ammonia possess strong reducing nature? |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/sodium-1215510" style="font-weight:bold;" target="_blank" title="Click to know more about SODIUM">SODIUM</a> atoms<br/>sodium <a href="https://interviewquestions.tuteehub.com/tag/hydride-1034032" style="font-weight:bold;" target="_blank" title="Click to know more about HYDRIDE">HYDRIDE</a><br/> sodium amide<br/><a href="https://interviewquestions.tuteehub.com/tag/solvated-7743184" style="font-weight:bold;" target="_blank" title="Click to know more about SOLVATED">SOLVATED</a> <a href="https://interviewquestions.tuteehub.com/tag/electrons-969138" style="font-weight:bold;" target="_blank" title="Click to know more about ELECTRONS">ELECTRONS</a>.</p>Solution :solvated electrons.</body></html> | |
| 49331. |
A solution of silver nitrate was stirred with iron rod will it cause any change in the concentration of silver and nitrate ions ? |
| Answer» <html><body><p></p>Solution :Since `<a href="https://interviewquestions.tuteehub.com/tag/e-444102" style="font-weight:bold;" target="_blank" title="Click to know more about E">E</a>^(@)` of `Fe^(2+)//Fe` (-0.44 V) is <a href="https://interviewquestions.tuteehub.com/tag/lower-1080637" style="font-weight:bold;" target="_blank" title="Click to know more about LOWER">LOWER</a> than that of `Ag^(+)//Ag(+0.80 V)` <a href="https://interviewquestions.tuteehub.com/tag/electrode-968538" style="font-weight:bold;" target="_blank" title="Click to know more about ELECTRODE">ELECTRODE</a> therefore `Ag^(+)` gets <a href="https://interviewquestions.tuteehub.com/tag/reduced-1181337" style="font-weight:bold;" target="_blank" title="Click to know more about REDUCED">REDUCED</a> and Fe gets oxidised As a result concentration of `Ag^(+)` ions <a href="https://interviewquestions.tuteehub.com/tag/decreases-946143" style="font-weight:bold;" target="_blank" title="Click to know more about DECREASES">DECREASES</a> while that of `NO_(3)^(-)` ions remains unchanged <br/> `2 Ag^(+) (aq)+Fe(s)rarr 2 Ag(s)+Fe^(2+)(aq)`</body></html> | |
| 49332. |
A solution of potassium hydroxide is used to absorb carbon evolved during the estimation of carbon in an organic compound. Explain. |
| Answer» <html><body><p></p>Solution :Carbon dioxide <a href="https://interviewquestions.tuteehub.com/tag/reacts-1178303" style="font-weight:bold;" target="_blank" title="Click to know more about REACTS">REACTS</a> with `KOH` present in the solution to form <a href="https://interviewquestions.tuteehub.com/tag/soluble-3046305" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUBLE">SOLUBLE</a> <a href="https://interviewquestions.tuteehub.com/tag/potassium-603438" style="font-weight:bold;" target="_blank" title="Click to know more about POTASSIUM">POTASSIUM</a> carbonate and can be <a href="https://interviewquestions.tuteehub.com/tag/estimated-975638" style="font-weight:bold;" target="_blank" title="Click to know more about ESTIMATED">ESTIMATED</a>. ltbr. `2KOH+CO_(2) rarr K_(2)CO_(3) +H_(2)O`</body></html> | |
| 49333. |
A solution of palmitic acid in benzine contains 4.24 g of acid per litre.When this solution is dropped on water surface, benzene gets evaporated and palmitic acids forms a unimolecular film on surface. If we wish to cover an area of 500 cm^(2) with unimolecular film, what volume of solution should be used? The area covered by one palmitic acid molecule may be taken as 0.21 nm^(2). Mol.wt.of palmitic acid is 256. |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :`2.386xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>)"<a href="https://interviewquestions.tuteehub.com/tag/litre-1075864" style="font-weight:bold;" target="_blank" title="Click to know more about LITRE">LITRE</a>"`;</body></html> | |
| 49334. |
A solution of Na_(2)S_(2)O_(3) is standardized iodometrically against 0.1262 g of KBrO_(3). This process required 45 mL of the Na_(2)S_(2)O_(3) solution. What is the strength of the Na_(2)S_(2)O_(3)? (K = 39, Br = 80) |
| Answer» <html><body><p>`0.2M`<br/>`0.1M`<br/>`0.05M`<br/>`0.1N`</p>Solution :M.eq. `Na_(2)S_(2)O_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)` = M.eq `I_(2)` <br/> `45xxMxx1=nxx2impliesn=(45M)/(2)...(1)` <br/> M.eq `I_(2)` = M.eq `KBrO_(3)` <br/> `nxx2=6xx(0.1262)/(<a href="https://interviewquestions.tuteehub.com/tag/167-1795173" style="font-weight:bold;" target="_blank" title="Click to know more about 167">167</a>)xx1000` <br/> `impliesn=(3xx0.1262xx1000)/(167)...(2)` <br/> From (1) & (2), `(45M)/(2)=(126.2xx3)/(167)impliesM=0.1`</body></html> | |
| 49335. |
A solution of Na_2S_2O_3 is standardised iodometrically by using K_2Cr_2O_7. The equivalent weight of K_2Cr_2O_7 in this method is: |
| Answer» <html><body><p>Mol. Wt/<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a><br/>Mol. Wt/<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a><br/>Mol. Wt/<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a><br/>equal to mol wt. </p>Solution :`Cr_(2)O_(<a href="https://interviewquestions.tuteehub.com/tag/7-332378" style="font-weight:bold;" target="_blank" title="Click to know more about 7">7</a>)^(2-) to 2Cr^(3+)` <a href="https://interviewquestions.tuteehub.com/tag/eq-446394" style="font-weight:bold;" target="_blank" title="Click to know more about EQ">EQ</a>. wt. `=("Mol wt")/6`</body></html> | |
| 49336. |
A solution of m-chloroaniline, m-chlorophenol, m-chlorobenzoic acid in ethyl acetate was extracted initially with a saturated solution of NaHCO_(3) to give fraction A, the leftover organic phase was extracted with dil. NaOH to give fraction B. The final organic layer was labelled as fraction C. Fractions A, B and C contains respectively. |
| Answer» <html><body><p>m-chlorobenzoic <a href="https://interviewquestions.tuteehub.com/tag/acid-847491" style="font-weight:bold;" target="_blank" title="Click to know more about ACID">ACID</a>, m-chlorophenol and m-chloroaniline<br/>m-chlorophenol, m-chlorobenzoic acid and m-chloroaniline<br/>m-chloroaniline, m-chlorophenol and m-chlorobenzoic acid<br/>m-chlorobenzoic acid, m-chloroaniline and m-chlorophenol</p>Solution :m-chlorobenzoic acid being the most <a href="https://interviewquestions.tuteehub.com/tag/acidic-847601" style="font-weight:bold;" target="_blank" title="Click to know more about ACIDIC">ACIDIC</a> can be separated by a weak base like `NaHCO_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)` and <a href="https://interviewquestions.tuteehub.com/tag/hence-484344" style="font-weight:bold;" target="_blank" title="Click to know more about HENCE">HENCE</a> will be labelled <a href="https://interviewquestions.tuteehub.com/tag/fraction-458259" style="font-weight:bold;" target="_blank" title="Click to know more about FRACTION">FRACTION</a> A. m-chlorophenol is not as acidic as m-chlorobenzoic acid, and can be separated by a stronger base like NaOH, and hence can be labelled as fraction B. m-chloroaniline being a base, does not react with either of the bases and hence would be labelled as fraction C</body></html> | |
| 49337. |
A solution of HCI is prepared by dissolving 7.30 g of hydrogen chloride gas in 100 ml of water. Find the molarity of the solution. If 50.0 cm^3 of this solution is treated with 3.50 g of zinc, what volume of H_2 measured at S.T.P. will be evolved ? |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :1.12 <a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a></body></html> | |
| 49338. |
A solution of H_(2)O_(2) is titrated against a solution of KMnO_(4). The reaction is : 2MnO_(4)^(-)+5H_(2)O_(2)+6H^(+) to 2Mn^(2+)+5O_(2)+8H_(2)O If it requires 46.9 mL of 0.145 M KMnO_(4) to oxidise 20 g of H_(2)O_(2), the mass percentage of H_(2)O_(2) in this solution is : |
| Answer» <html><body><p>2.9<br/>29<br/>21<br/><a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>.9</p>Solution :Number of moles of `KMnO_(4)=(MV)/(<a href="https://interviewquestions.tuteehub.com/tag/1000-265236" style="font-weight:bold;" target="_blank" title="Click to know more about 1000">1000</a>)=(0.145xx46.9)/(1000)` <br/> `=6.8xx10^(-3)` <br/> Number of moles of `H_(2)O_(2)=6.8xx10^(-3)xx2.5=0.017` <br/> Mass of `H_(2)O_(2)=0.017xx34=0.578` <br/> Mass % of `H_(2)O_(2)=(0.578)/(<a href="https://interviewquestions.tuteehub.com/tag/20-287209" style="font-weight:bold;" target="_blank" title="Click to know more about 20">20</a>)xx100=2.9`</body></html> | |
| 49339. |
A solution of glycine hydrochloride contains the chloride ion and the glycinium ion, .^(+)NH_(3) - CH_(2) - COOH, which is a diprotic acid, H_(3)N^(+) -CH_(2) - COOH + H_(2)O hArr H_(3)N^(+) - CH_(2) - COO^(-) + H_(3)O^(+), K_(1) = 4.47 x10^(-3) H_(3)N^(+) - CN_(2) - COO^(-) + H_(2)O hArr H_(2)N - CH_(2) - COO^(-) + H_(3)O^(+), K_(2) = 1.66x10^(-10) Calculate pH of a 0.05 M of glycine hydrochloride. |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>.89`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>.52`<br/>`8.91`<br/>`9.18`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`(<a href="https://interviewquestions.tuteehub.com/tag/ph-1145128" style="font-weight:bold;" target="_blank" title="Click to know more about PH">PH</a> =pKa_(1) + pKa_(2))/(2)`</body></html> | |
| 49340. |
A solution of ferrous oxalate has been prepared by dissoliving 3.6 g L^(-1) calculate the volume of 0.01 M KMnO_(4) solution required for complete oxidatin of 100 mL of lferrous oxalte soluton in acidic medium |
| Answer» <html><body><p><br/></p>Solution :Molarity of `FeC_(2)O_(4)` solution `=("mass in" gL^(-<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>))/("<a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a> mass in" <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a> mol^(-1))=(3.6)/(144)=0.025M` <br/> The balanced chemical equation for the redox reaction is : <br/> `5FeC_(2)O_(4)+3MinO_(4)^(2-)+24H^(+)rarr5Fe^(3+)+<a href="https://interviewquestions.tuteehub.com/tag/3mn-1863569" style="font-weight:bold;" target="_blank" title="Click to know more about 3MN">3MN</a>^(2+)10CO_(2)+12H_(2)O` <br/> Applying molarity equation we have <br/> `(0.01xxV)/(3)(MnO_(4)^(-))=(0.025xx100)/(5)(FeC_(2)O_(4)) "in" V=(0.025xx100xx3)/(5xx0.01)=150 mL`</body></html> | |
| 49341. |
A solution of ethanol in water is 1.6 molal. How many grams of ethanol are present in 500 g of the solution ? |
| Answer» <html><body><p></p>Solution :1.6 m solution would contain 1.6 <a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a> of <a href="https://interviewquestions.tuteehub.com/tag/ethanol-975823" style="font-weight:bold;" target="_blank" title="Click to know more about ETHANOL">ETHANOL</a> dissolved in 1000 g of water. <br/> Molecular mass of ethanol `(C_(2)H_(5)OH) = 46` <br/> Mass of ethanol dissolved = `46 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 1.6 = 73.6` <br/>g Total mass of solution = `73.6 + 1000 = 1073.6` g<br/> <a href="https://interviewquestions.tuteehub.com/tag/thus-2307358" style="font-weight:bold;" target="_blank" title="Click to know more about THUS">THUS</a>, 1073.6 g of solution contain 73.6 g of ethanol. Hence, the mass of ethanol present in 500 g of the solution `=73.6/(1073.6)xx 500 = 34.28 g`</body></html> | |
| 49342. |
A solution of colourless salt H on boiling with excess NaOH produces a non-flammable gas. The gas evolution ceases after sometime. Upon addition of Zn dust to the same solution, the gas evolution restarts. The colourless salt(s) H is (are) |
| Answer» <html><body><p>`NH_(4)NO_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)`<br/>`NH_(4)NO_(2)`<br/>`NH_(4)<a href="https://interviewquestions.tuteehub.com/tag/cl-408888" style="font-weight:bold;" target="_blank" title="Click to know more about CL">CL</a>`<br/>`(NH_(4))_(2)SO_(4)`</p>Solution :The colourless salt H may be either `NH_(4)NO_(3)orNH_(4)NO_(2)` and the non-inflammable gas is `NH_(3)`<br/> (a) `NH_(4)NO_(3)+NaOHoverset(Delta)toNH_(3)+NaNO_(3)+H_(2)O` <br/> Non-inflammable gas <br/> `4Zn+7NaOH+NaNO_(3)overset(Delta)to4Na_(2)ZnO_(2)+NH_(3)+2H_(2)O` <br/> Sod. zincate Gas <a href="https://interviewquestions.tuteehub.com/tag/evolution-977684" style="font-weight:bold;" target="_blank" title="Click to know more about EVOLUTION">EVOLUTION</a> restarts <br/> (b) `NH_(4)NO_(2)+NaOHoverset(Delta)toNH_(3)+NaNO_(2)+H_(2)O` <br/> `3Zn+5NaOH+NaNO_(2)overset(Delta)to3Na_(2)ZnO_(2)+NH_(3)+H_(2)O`</body></html> | |
| 49343. |
A solution of colourless salt H on boiling with excess NaOH produces a non-flammable gas. The gas evolution ceases after some time. Upon addition of Zn dust to the same solution, the gas evolution restarts. The colourless salt(s) H is (are): |
| Answer» <html><body><p>`NH_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)NO_(3)`<br/>`NH_(4)NO_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)`<br/>`NH_(4)<a href="https://interviewquestions.tuteehub.com/tag/ci-408488" style="font-weight:bold;" target="_blank" title="Click to know more about CI">CI</a>`<br/>`(NH_(4))_(2)SO_(4)`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`NH_(4)^(+)` salts liberate `NH_(3)` with `NaOH` <br/> `NH_(4)NO_(3)` liberates `N_(2)O` (<a href="https://interviewquestions.tuteehub.com/tag/flammable-2646523" style="font-weight:bold;" target="_blank" title="Click to know more about FLAMMABLE">FLAMMABLE</a> gas)</body></html> | |
| 49344. |
A solution of CoCI_(2).6H_(2)O in isopropyl alcohol and water is purple. The color change to blue when we add |
| Answer» <html><body><p>concentrated HCI<br/>`AgNO_(3)(aq.)<br/>both (1) and (2)<br/>none of these</p>Solution :The purple color is due to the <a href="https://interviewquestions.tuteehub.com/tag/mixture-1098735" style="font-weight:bold;" target="_blank" title="Click to know more about MIXTURE">MIXTURE</a> of `[Co(H_(2)O)_(6)]^(2+)` (pink) and `[CoCl_(4)]^(2-)` (blue): <br/> `[Co(OH_(2))_(6)]^(2+)+4Cl^(-)hArr[CoCl_(4)]^(2-)+6H_(2)O` <br/> When we add concentrated HCI, excess `Cl^(-)` shifts the equilibrium to the right (blue). <a href="https://interviewquestions.tuteehub.com/tag/adding-2399902" style="font-weight:bold;" target="_blank" title="Click to know more about ADDING">ADDING</a> `AgNO_(3)` (aq.) removes some `Cl^(-)` by <a href="https://interviewquestions.tuteehub.com/tag/precipitation-16670" style="font-weight:bold;" target="_blank" title="Click to know more about PRECIPITATION">PRECIPITATION</a> of `AgCl(s)` and favours the reaction to the left (<a href="https://interviewquestions.tuteehub.com/tag/produces-1167808" style="font-weight:bold;" target="_blank" title="Click to know more about PRODUCES">PRODUCES</a> more `[Co(OH_(2))_(6)]^(2+)` and the resulting solution is pink.</body></html> | |
| 49345. |
A solution of a non-volatile solute in water has a boiling point of 375.3K. Calculate the vapour pressure of water above this solution at 338K. Given, p_(0) (water) = 0.2467 atm at 338K and K_(b) for water = 0.52. |
| Answer» <html><body><p></p>Solution :`DeltaT_(b)=(375.3-373.16)=2.15K` <br/> We know that, <br/> `DeltaT_(b)`= Molality `<a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> K_(b)` <br/> 2.15 = Molality `xx 0.52` <br/> Molality `= (2.15)/(0.52) = 4.135` <br/> i.e., 4.135 moles of the <a href="https://interviewquestions.tuteehub.com/tag/solute-1217068" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTE">SOLUTE</a> present in 1000g of water (55.5 moles) <br/> Mole fraction of water `= (55.5)/(4.135+55.5)=(55.5)/(59.635)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/vapour-1442886" style="font-weight:bold;" target="_blank" title="Click to know more about VAPOUR">VAPOUR</a> pressure of water above solution <br/> = Mole fraction `xx p_(0)` <br/> `=(55.5)/(59.635)xx0.2467=0.23`atm.</body></html> | |
| 49346. |
A solution of a metal ion when treated with KI solution gives a red precipitate which dissolves in excess of KI solution to give a colourless solution. Moreover, the solution of the same metal ion on treatment with the solution of cobalt (II) thiocyanate gives rise to a deep blue crystalline precipitate. The metal ion is |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/pb-597752" style="font-weight:bold;" target="_blank" title="Click to know more about PB">PB</a>^(2+)`<br/>`Hg^(2+)`<br/>`Cu^(2+)`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/co-920124" style="font-weight:bold;" target="_blank" title="Click to know more about CO">CO</a>^(2+)`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :It is the correct answer : <br/>`{:("Hg"^(2+),+,2l^(-),<a href="https://interviewquestions.tuteehub.com/tag/rarr-1175461" style="font-weight:bold;" target="_blank" title="Click to know more about RARR">RARR</a>,Hgl_(2),),(,,(Kl),,("Scarlet red ppt"),),(Hgl_(2),+,2Kl,rarr,underset(("Colourless"))(K_(2)Hgl_(4)),),("Hg"^(2+),+,Co(<a href="https://interviewquestions.tuteehub.com/tag/scn-2255678" style="font-weight:bold;" target="_blank" title="Click to know more about SCN">SCN</a>)_(2),rarr,underset(("Deep blue crystalline ppt."))(Co [Hg (SCN)_(2)]),):}`</body></html> | |
| 49347. |
A solution of a compound X in dilute HCl on treatment with a solution of BaCl_(2) gives a white precipitate of compound Y which is insoluble in conc. HNO_(3) and and conc. HCl. Compound X imparts golden yellow colour to the flame. underset(("immparts golden yellow colour"))(X("Solution in dilute HCl"))+BaCl_(2) to underset("White")(Y)underset("Conc. HCl")overset("Conc. "HNO_(3))toInsoluble What are compounds X and Y? |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a> is `MgCl_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)` and Y is `BaSO_(4)`<br/>X is `CaCl_(2) and Y` is `BaSO_(4)`<br/>X is `Na_(2)SO_(4)` and Y is `BaSO_(4)`<br/>X is `MgSO_(4)` and Y is `BaSO_(4)`</p>Solution :`underset((X))(Na_(2)SO_(4))+BaCl_(2) to underset((Y))(BaSO_(4))+2NaCl`</body></html> | |
| 49348. |
A solution of a colourles salt H on boiling with excess of NaOH produces a non-flammable gas and the evolution of gas ceases after sometime. Upon addition of Zn dust to the same solution, the evolution of gas restarts. The colourless salt (S) 'H' is (are) |
| Answer» <html><body><p>`NH_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)NO_(3)`<br/>`NH_(4)NO_(2)`<br/>`NH_(4)Cl`<br/>`(NH_(4))_(2)SO_(4)`</p>Solution :The <a href="https://interviewquestions.tuteehub.com/tag/colourless-422296" style="font-weight:bold;" target="_blank" title="Click to know more about COLOURLESS">COLOURLESS</a> <a href="https://interviewquestions.tuteehub.com/tag/salts-1193872" style="font-weight:bold;" target="_blank" title="Click to know more about SALTS">SALTS</a> are `NH_(4)NO_(3)` (a) and `NH_(4)NO_(2)` (<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>). <br/> `underset((a))(Na_(4)NO_(3)) + NaOH rarr NaNO_(3) + underset(("Colourless gass"))(NH_(3) + H_(2)O)` <br/> `NaNO_(3) + <a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a>[H] overset((Zn - "dust"))rarr NaOH + NH_(3) + 2H_(2)O` <br/> `underset((b))(NH_(4)NO_(2)) + NaOH rarr NaNO_(2) + NH_(3) + H_(2)O` <br/> `NaNO_(2) + 6[H] overset((Zn - "dust"))rarr NaOH + NH_(3) + H_(2)O`</body></html> | |
| 49349. |
A solution of 10 mL of (M)/(10) FeSO_(4) was titrated with KMnO_(4) solution in acidic medium, the amount of KMnO_(4) used will be : |
| Answer» <html><body><p>10 mL of 0.5 M<br/>10 mL of 0.<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> M<br/>10 mL of 0.02 M<br/>5 mL of 0.1 M</p>Solution :The <a href="https://interviewquestions.tuteehub.com/tag/involved-7257329" style="font-weight:bold;" target="_blank" title="Click to know more about INVOLVED">INVOLVED</a> reaction is : <br/> `2KMnO_(4)+8H_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)SO_(4)+10FeSO_(4) to 5Fe_(2)(SO_(4))_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)+2MnSO_(4)+K_(2)SO_(4)+8H_(2)O` <br/> `(M_(1)V_(1))/(2)=(0.1xx10)/(10)` <br/> `M_(1)V_(1)=0.2` which is possible in (c )</body></html> | |
| 49350. |
A solution of 1 molal concentration of a solute will have maximum boiling point elevation when the solvent is : |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/ethyl-976035" style="font-weight:bold;" target="_blank" title="Click to know more about ETHYL">ETHYL</a> alcohol<br/>acetone<br/>benzene<br/>chloroform</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |