Explore topic-wise InterviewSolutions in Class 11.

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49451.

(A) : Silica is used as acidic flux in metallurgy (R ) : Silica is basic in nature

Answer» <html><body><p>A and R are true, R explains A <br/>A and R are true, R does not <a href="https://interviewquestions.tuteehub.com/tag/explain-447165" style="font-weight:bold;" target="_blank" title="Click to know more about EXPLAIN">EXPLAIN</a> 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> :<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>
Discussion
49452.

A siliate used in talcum powder

Answer» <html><body><p>Consists of chain which are very long<br/>is known as <a href="https://interviewquestions.tuteehub.com/tag/talc-1238597" style="font-weight:bold;" target="_blank" title="Click to know more about TALC">TALC</a> and is a pure magnesium <a href="https://interviewquestions.tuteehub.com/tag/silicate-1207627" style="font-weight:bold;" target="_blank" title="Click to know more about SILICATE">SILICATE</a> of the form 3MgO `4SiO_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>). H_(2)`<br/>is a three dimensional silicae<br/>is a sheet silicate</p>Solution :Pure magnesium silicate</body></html>
Discussion
49453.

A sigma-bonded molecule MX_(3) is T-shaped. The number of non-bonding pairs of electrons is-

Answer» <html><body><p>0<br/>2<br/>1<br/>3</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>
Discussion
49454.

(a) Show the formation of the electrophile in the following reactions : (i) CI_(2)+AICI_(3), (ii) HNO_(3)+H_(2)SO_(4), (iii) Br_(2)+Fe, (iv) H_(2)SO_(4), (v) H_(2)S_(2)O_(7), Fuming sulphuric acid. (b) How do substituent groups on an aromatic rign influence the course of electrophilic aromatic substituion ? Classify them by their effects.

Answer» <html><body><p></p>Solution :(a) (i) `CI_(2)+AICI_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)toCI^(+)+AICI_(4)^(-)` <br/> (ii) `HONO_(2)+HOSO_(3)HtoH_(2)O+NO_(2)^(+)+HSO_(4)^(-)` <br/>(iii) `2Fe+3FeBr_(2)to2FeBr_(3)` <br/> `FeBr_(3)+Br_(2)toBr^(+)+FeBr_(4)^(-)` <br/> (iv) `2H_(2)SO_(4)toH_(3)O^(+)+HSO_(4)^(-)+SO_(3)` <br/> (v)`H_(2)S_(2)O_(7)toH^(+)+HSO_(4)^(+)+SO_(3)` <br/> (<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>) The substituents on an aromatic ring affect the electrophilic substitution in two ways : <br/> 1. Reactivity : The compound is more reactive than benzene, the groups present are activating. In case, the compound is less reactive than benzene, the groups present are deactivating. <br/>2. Orientation : Whether electrophile (E) enters ortho, para or meta. There are three <a href="https://interviewquestions.tuteehub.com/tag/classes-16111" style="font-weight:bold;" target="_blank" title="Click to know more about CLASSES">CLASSES</a> of <a href="https://interviewquestions.tuteehub.com/tag/substituent-1231626" style="font-weight:bold;" target="_blank" title="Click to know more about SUBSTITUENT">SUBSTITUENT</a> groups: <br/> (i)All activaing groups direct E to ortho or parapositions. <br/> (ii) Most deactivating groups direct E to meta-positions. ltbegt (iii) A few deactivating groups, e.g., halogens direct E to ortho or para-positions.</body></html>
Discussion
49455.

(A): SiF_4is non polar even though fluorine is much more electronegative than silicon (B) : The four bond dipoles cancel one another in SiF_4molecule

Answer» <html><body><p>Both (A) and (R) are <a href="https://interviewquestions.tuteehub.com/tag/true-713260" style="font-weight:bold;" target="_blank" title="Click to know more about TRUE">TRUE</a> and (R) is the correct explanation of (A)<br/>Both (A) and (R) are true and (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> :A</body></html>
Discussion
49456.

A series of lines in the spectrum of atomic hydrogen lies at 656.46 n, 486.27 nm, 439.17 nm and 410.29 nm. What is the wavelength of the next line in this series? What is the ionisation energy of the atom when it is ini the lower state of transition?

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> :`lamda_(<a href="https://interviewquestions.tuteehub.com/tag/next-578185" style="font-weight:bold;" target="_blank" title="Click to know more about NEXT">NEXT</a>)=397.15nm,IE=3.240`<a href="https://interviewquestions.tuteehub.com/tag/ev-976207" style="font-weight:bold;" target="_blank" title="Click to know more about EV">EV</a></body></html>
Discussion
49457.

A semiconductor of Ge can be made p - type by adding

Answer» <html><body><p>Trivalent impurity <br/>Tetravalent<br/>Pentavalent <br/>Divalent impurity </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> :p-type semiconductors are made by <a href="https://interviewquestions.tuteehub.com/tag/adding-2399902" style="font-weight:bold;" target="_blank" title="Click to know more about ADDING">ADDING</a> impurity of <a href="https://interviewquestions.tuteehub.com/tag/previous-592857" style="font-weight:bold;" target="_blank" title="Click to know more about PREVIOUS">PREVIOUS</a> group elements. For making p-type semiconductor of Ge (`14^(th)` group), it is to be doped with trivalent impurity.</body></html>
Discussion
49458.

A semiconductor of Ge can be made p-type by adding

Answer» <html><body><p>rtivalent <a href="https://interviewquestions.tuteehub.com/tag/impurity-1039200" style="font-weight:bold;" target="_blank" title="Click to know more about IMPURITY">IMPURITY</a><br/>tetravalent impurity<br/>pentavalent impurity<br/>divalent impurity</p>Solution :Ge is a <a href="https://interviewquestions.tuteehub.com/tag/group-1013370" style="font-weight:bold;" target="_blank" title="Click to know more about GROUP">GROUP</a> <a href="https://interviewquestions.tuteehub.com/tag/14-272882" style="font-weight:bold;" target="_blank" title="Click to know more about 14">14</a> element. Positive holes can be <a href="https://interviewquestions.tuteehub.com/tag/created-7257327" style="font-weight:bold;" target="_blank" title="Click to know more about CREATED">CREATED</a> by adding group 13elementi.e., trivalent impurity.</body></html>
Discussion
49459.

A secondary pollulant is

Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/co-920124" style="font-weight:bold;" target="_blank" title="Click to know more about CO">CO</a><br/>`CO_2`<br/><a href="https://interviewquestions.tuteehub.com/tag/pan-590075" style="font-weight:bold;" target="_blank" title="Click to know more about PAN">PAN</a><br/>Aerosol</p>Solution :PAN</body></html>
Discussion
49460.

A sealed container was filled with 1 mol of A_2(g), 1 mol B_2(g) at 800 K and total pressure 1.00 bar. Calculate the amounts of the components in the mixture at equilibrium given that K = 1 for the reaction : A_2(g)+ B_2(g)hArr 2AB(g)

Answer» <html><body><p></p>Solution :`A_2(g) + B_2(g) hArr 2AB(g)`<br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/FM_CHE_XI_V02_C08_E01_048_S01.png" width="80%"/><br/> Total no. of moles `= <a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>-x+ 1 - x + 2x=2` <br/> `K_P = ((P_(<a href="https://interviewquestions.tuteehub.com/tag/ab-360636" style="font-weight:bold;" target="_blank" title="Click to know more about AB">AB</a>))^2)/((P_(A_2))(P_(B_2))) =(((2x)/2xxP)^2)/((((1-x))/2xxP)(((1-x))/2xxP))` <br/> `K_P = (4x^2)/((1-x)^2)`<br/> Given that `K_P = (4x^2)/((1-x)^2) = 1` <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> 4x^2 = (1-x)^2 rArr 4x^2 =1 +x^2 -2x`<br/> `3x^2 + 2x -1 = 0`<br/> `x = (-2 pmsqrt(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a> - 4 xx 3 xx (-1)))/(2(3))` <br/> `x = (-2 pmsqrt(4 + <a href="https://interviewquestions.tuteehub.com/tag/12-269062" style="font-weight:bold;" target="_blank" title="Click to know more about 12">12</a>))/6 = (-2pm sqrt16)/6 = (-2+4)/6,(-2-4)/6 = 2/6, (-6)/6` <br/> ` x = 0.33 , -1` (not posoible) <br/> `:.[A_2]_(eq) = 1 - x =1 - 0.33 = 0.67` <br/> `[B_2]_(eq) = 1- x = 1-0.33 = 0.67`<br/> `[AB]_(eq) = 2x = 2 xx 0.33 = 0.66`</body></html>
Discussion
49461.

A sealed container was filled with 1 mol of A_(2)(g) 1 mol B_(2)(g) at 800 K and total pressure 1.00 bar. Calculate the amounts of the components in the mixture at equilibrium given that K=1 for the reaction A_(2)(g)+B_(2)(g)hArr2AB(g)

Answer» <html><body><p></p>Solution :`A_(2)(g)+B_(2)(g)hArr2AB(g)` <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/SUR_CHE_XI_V02_C08_E01_050_S01.png" width="80%"/> <br/> <a href="https://interviewquestions.tuteehub.com/tag/total-711110" style="font-weight:bold;" target="_blank" title="Click to know more about TOTAL">TOTAL</a> no. of moles `=1-x+1-x+<a href="https://interviewquestions.tuteehub.com/tag/2x-301182" style="font-weight:bold;" target="_blank" title="Click to know more about 2X">2X</a>=2` <br/> `K_(P)=((P_(AB))^(2))/((P_(A_(2)))(P_(B_(2))))=(((2x)/(2)xxP)^(2))/((((1-x))/(2)xxP)((1-x)/(2)xxP))` <br/> `K_(P)=(<a href="https://interviewquestions.tuteehub.com/tag/4x-319255" style="font-weight:bold;" target="_blank" title="Click to know more about 4X">4X</a>^(2))/((1-x)^(2))` <br/> Given that `K_(P)=1,(4x^(2))/((1-x)^(2))=1` <br/> `rArr4x^(2)=(1-x)^(2)` <br/> `rArr4x^(2)=1+x^(2)-2x` <br/> `3x^(2)+2x-1=0` <br/> `x=(-2pmsqrt(4-(4xx3xx-1)))/(2(3))` <br/> `x=(-2pmsqrt(4+12))/(<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/> `=(-2pmsqrt(16))/(6)` <br/> `=(2)/(6),(-2-4)/(6)` <br/> `x=0.33,-1("not <a href="https://interviewquestions.tuteehub.com/tag/possible-592355" style="font-weight:bold;" target="_blank" title="Click to know more about POSSIBLE">POSSIBLE</a>")` <br/> `:.[A_(2)]_(eq)=1-x=1-0.33=0.67` <br/> `[B_(2)]_(eq)=1-x=1-0.33=0.67` <br/> `[AB]_(eq)=2x=2xx0.33=0.66`.</body></html>
Discussion
49462.

A sea diver at depth of 45m exhales a bubble of air that is 1.0 cm in radius. Assuming the ideal behaviour, find out radius of this bubble as it breaks at the surface of water?

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>.75 cm <br/>1.50 cm <br/>1.25 cm <br/>0.75 cm </p>Solution :Pressure of <a href="https://interviewquestions.tuteehub.com/tag/45-316951" style="font-weight:bold;" target="_blank" title="Click to know more about 45">45</a> m <a href="https://interviewquestions.tuteehub.com/tag/water-1449333" style="font-weight:bold;" target="_blank" title="Click to know more about WATER">WATER</a> in terms of mercury level : <br/> `h_1d_1 = h_2d_2` <br/>`45 xx 1 = h_2 xx 13.6 = <a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a> = 3.3 m`. <br/> `:.` Pressure at 45 m depth = `(0.76 + 3.3) = 4.06 m.` <br/> `P_1V_1 = P_2V_2` <br/> `4.06 xx 4/3 pi (1)^3 = 0.76 xx 4/3 pi (r_(cm))^(3)` <br/> `implies r_(cm) root(3)((4.06)/(0.76)) , r_(cm) = 1.75 cm`</body></html>
Discussion
49463.

A scientist attempts to replace a few carbon atoms in 1.0 g of diamond with boron atoms or nitrogen atoms in separate experiments. Which of the following is correct ?

Answer» <html><body><p>The resulting material with B droping will be an <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a>-type semiconductor<br/>The resulting material with B droping will be an p-type semiconductor<br/>B droping is NOT possible as B cannot from <a href="https://interviewquestions.tuteehub.com/tag/multiple-1105557" style="font-weight:bold;" target="_blank" title="Click to know more about MULTIPLE">MULTIPLE</a> bonds<br/>the resulting material with N droping will be a p-type semiconductore</p>Answer :B</body></html>
Discussion
49464.

(A) Schrodinger wave equation (B) Heisenberg uncertainty principle (C) Millikan oil drop experiment (D) Chadwick.s discovery of neutron Of the scientific milestones listed above, which is the most ancient and most modern respectively ?

Answer» <html><body><p>A, D<br/>C, D<br/><a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>, A<br/>C, A</p>Answer :B</body></html>
Discussion
49465.

A saturated solution of sparingly soluble lead chloride on analysis was found to contain 11.84 g/ litre of the salt at room temperature. Calculate the solubility product constantat room temperature. (At. wt . : Pb= 207, Cl = 35.5 )

Answer» <html><body><p><br/></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/solubility-1217018" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUBILITY">SOLUBILITY</a> of `PbCl_(2) = (11.84)/(207+2xx35.5) "<a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a>" L^(-1) = 4.259 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 10^(-2) "mol " L^(-1)` <br/> `PbCl_(2) rarr Pb^(2+)+2 <a href="https://interviewquestions.tuteehub.com/tag/cl-408888" style="font-weight:bold;" target="_blank" title="Click to know more about CL">CL</a>^(-), K_(sp)=[Pb^(2+)][Cl^(-)]^(2) = (4.259xx10^(-2))(2xx4.259xx10^(-2))^(2) = 3.09xx10^(-4)`</body></html>
Discussion
49466.

A saturated solution of XCl_(3) has a vapour pressure 17.20 mm Hg at 20^(@)C, while pure water vapour pressure is 17.25 mm Hg. Solubilityproduct (K_(sp)) of XCl_(3) at 20^(@)C is :

Answer» <html><body><p>`9.8 xx <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/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)`<br/>`10^(-5)`<br/>2.56x`10^(-<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/><a href="https://interviewquestions.tuteehub.com/tag/7x-336309" style="font-weight:bold;" target="_blank" title="Click to know more about 7X">7X</a>`10^(-5)`</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>
Discussion
49467.

A saturated solution of o- nitrophenol has a pH equal to 4.53 then its solubility in water is (pK_a=7.23)

Answer» <html><body><p>` 2.085 g//"lit"` <br/>` 20. 85 g //"lit"` <br/>` 10. 425g//"lit"` <br/>` 1.0425 g//"lit"` </p>Solution :` O- " nitrophenol" _(s) hArr O- " nitro <a href="https://interviewquestions.tuteehub.com/tag/phenol-1152951" style="font-weight:bold;" target="_blank" title="Click to know more about PHENOL">PHENOL</a> "_((aq) ) ` <br/> ` O- "nitrophenol"_((aq)) hArr O- " nitro phenol " + H^(+) ` <br/> ` S(1-<a href="https://interviewquestions.tuteehub.com/tag/alpha-858274" style="font-weight:bold;" target="_blank" title="Click to know more about ALPHA">ALPHA</a> )"" S alpha "" S alpha ` <br/> ` [H^(+) ]=S alpha =sqrt( SKa) ` <br/> ` -log [H^(+) ] =(-log S -log Ka)/( 2) ` <br/> ` <a href="https://interviewquestions.tuteehub.com/tag/ph-1145128" style="font-weight:bold;" target="_blank" title="Click to know more about PH">PH</a> =(-log S - log Ka )/( 2) ` <br/> ` pH =(- log S +<a href="https://interviewquestions.tuteehub.com/tag/pka-1145194" style="font-weight:bold;" target="_blank" title="Click to know more about PKA">PKA</a>)/( 2) `<br/> ` - log S = 2xx 4.53 -7.23 =1.83` <br/> ` rArr S = 1 .48 xx 10 ^(-2)"mol" //"lit" ` <br/> ` M. wt = 139 rArr S =2.06 g//lit `</body></html>
Discussion
49468.

A saturated solution of iodine in water contain 0.330g I_(2) per litre. More than this I_(2) can be dissolved in KI solution because of the following equilibrium. I_(2(g))+I^(-)hArrI_(3)^(-) A 0.100 M KI solution actually dissolves 12.5 g "iodine per litre", most of which is converted to I_(3)^(-). Assuming that the concentration of I_(2) in all saturated solution is the same, calculate the equilibrium constant for the above reaction. What is the effect of adding water to a clear saturated solution of I_(2) in the KI solution?

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> :`707 <a href="https://interviewquestions.tuteehub.com/tag/litre-1075864" style="font-weight:bold;" target="_blank" title="Click to know more about LITRE">LITRE</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>^(-<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)` ,</body></html>
Discussion
49469.

A saturated solution of H_2S in water has concetration of approximately 0.10 M. What is the pH of this soluton and equilibrium concentrations of H_2S,HS^(-) andS^(2-)? Hydrogen sulphide is a diprotic acid and its dissociation constants are K_(a_1)=9.1xx10^(-8) ,K_(a_2)=1.3xx10^(-3) "mol L"^(-1) respectively.

Answer» <html><body><p><br/></p>Solution :The <a href="https://interviewquestions.tuteehub.com/tag/second-1197322" style="font-weight:bold;" target="_blank" title="Click to know more about SECOND">SECOND</a> dissociation <a href="https://interviewquestions.tuteehub.com/tag/constant-930172" style="font-weight:bold;" target="_blank" title="Click to know more about CONSTANT">CONSTANT</a> is very very small than first dissociationconstant and therefore , the concentration of `H_3O^+` is obtained only from the first dissociation : <br/> `{:(,H_2S+H_2O hArr, H_3O^(+)+, <a href="https://interviewquestions.tuteehub.com/tag/hs-492079" style="font-weight:bold;" target="_blank" title="Click to know more about HS">HS</a>^-),("Initial conc.", 0.10,0,0),("At equilibrium" , 0.10-x, x ,x):}` <br/>(x is amount of `H_2S` dissociated ) <br/>`K_(a_1)=([H_3O^+][HS^-])/([H_2S])=9.1xx10^(-8)` <br/>or `(x xx x)/(0.10-x)=9.1xx10^(-8)` <br/> Solving `x=9.5x10^(-5)` <br/>`therefore [H_3O^+]=[HS^-]=9.5xx10^(-5) "mol L"^(-1)` <br/>`pH=-log [H_3O^+]=-log (9.5xx10^(-5))` <br/> `=-log 9.5 +5 =-0.98+5=4.02` <br/>`[H_2S]=0.10-9.5xx10^(-5)`=0.10 M <br/> To calculate the concentration of `S^(2-)` ion , we are to consider the second dissociation : <br/> `{:(,HS^(-)+H_2O hArr, H_3O^(+)+,S^(2-)),("Initial conc.", 9.5xx10^(-5), 9.5xx10^(-5),0),("At equi.",9.5xx10^(-5)-x, 9.5xx10^(-5)+x ,x):}`<br/>`K_(a_2)=([H_3O^+][S^(2-)])/([HS^-])` <br/> `=((9.5xx10^(-5)+x)xx x)/(9.5xx10^(-5)-x)` <br/> Assuming x to be very very small : <br/> `9.5xx10^(-5)-x <a href="https://interviewquestions.tuteehub.com/tag/approx-882876" style="font-weight:bold;" target="_blank" title="Click to know more about APPROX">APPROX</a> 9.5 xx10^(-5)` and `9.5xx10^(-5)+x` <br/>=`9.5xx10^(-5)` <br/>Solving `x=1.3xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/13-271882" style="font-weight:bold;" target="_blank" title="Click to know more about 13">13</a>)` <br/>`therefore [S^(2-)]=1.3xx10^(-13)` <br/> Thus, `[H_3O^+]=9.5xx10^(-5)` M , `[HS^-]=9.5xx10^(-5)` M, `[S^(2-)]=1.3xx10^(-13)M , [H_2S]`=0.01 M , pH=4.02</body></html>
Discussion
49470.

A saturated solution of Ca_3 (PO_4) _2has [Ca^(+2) ]=2xx 10 ^(-8)M and [PO_4^(-3) ] =1.6 xx 10 ^(-5) M , K_(sp)"of "Ca_3(PO_4) _2 is

Answer» <html><body><p>`3.2 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 10^(_13) ` <br/>` 3. 2 xx 10^(-34) ` <br/>` 2.048 xx 10^(-33)` <br/>none of these </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> :`Ca_3 (PO_4)hArr 3 Ca^(+2)+2PO_4^(_3)`<br/> ` K_(Sp)=[Ca^(+2)]^(3)[PO_4^(-3)]^(2) ` <br/> ` K_(Sp)=(<a href="https://interviewquestions.tuteehub.com/tag/2xx-1840186" style="font-weight:bold;" target="_blank" title="Click to know more about 2XX">2XX</a> 10 ^(-8))^(3) (1.6 xx 10^(-5) ) ^(2) ` <br/> ` = 8 xx 10^(-24)xx 2.56 xx 10 ^(-10)=2.048 xx 10 ^(-33)`</body></html>
Discussion
49471.

A saturated solution is prepared at 70^(circ)C containing 32.0g CusO_(4).5H_(2)O per 100 g solution. A 335 g sample of this solution is then cooled to 0^(circ)C so that. CuSO_(4).5H_(2)O crystallises out. If the concentration of a saturated solution at 0^(circ)C is 12.5g CuSO_(4).5H_(2)O per 100.0 g solution, how much of CuSO_(4).5H_(2)O is crystallised?

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> :`74.47 <a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a>`;</body></html>
Discussion
49472.

A saturated solution in AgX(K_(sp)=3xx10^(-12))and AgY(K_(sp)=10^(-12)) has conductivity 0.4xx10^(-6)Omega^(-1)cm^(-1). Given: Limiting molar conductivity of Ag^+=60Omega^(-1)cm^2mol^(-1) Limiting molar conductivity of X^(-)=90Omega^(-1)cm^2mol^(-1) The limiting molar conductivity of Y^(-) is (in Omega^(-1)cm^2mol^(-1)):

Answer» <html><body><p>290<br/>2900<br/>2.9<br/>None of these </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</body></html>
Discussion
49473.

A Saturated polyhalogen compound (A) on heating with zinc gives 2-Butyne. What should be the minimum number of halogen in one molecule of the reactant (A) to give the product.

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> :<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a></body></html>
Discussion
49474.

A saturated hydrocarbon has the formula C_(n)H_(12). The value of 'n' in this compound is

Answer» <html><body><p>4<br/>5<br/>6<br/>2</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>
Discussion
49475.

A Satellite is in an elliptic orbit around the earth with aphelion of 6R and perihelion of 2R

Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/position-1159826" style="font-weight:bold;" target="_blank" title="Click to know more about POSITION">POSITION</a> isomerism <br/>Chain isomerism <br/><a href="https://interviewquestions.tuteehub.com/tag/tautomerism-13486" style="font-weight:bold;" target="_blank" title="Click to know more about TAUTOMERISM">TAUTOMERISM</a> <br/>Geometrical isomerism </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> :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_TRG_AO_CHE_XI_V01_D_C02_E02_001_S01.png" width="80%"/><br/>They are position <a href="https://interviewquestions.tuteehub.com/tag/isomers-1052476" style="font-weight:bold;" target="_blank" title="Click to know more about ISOMERS">ISOMERS</a></body></html>
Discussion
49476.

A saturated 0.01 M H_(2) S solution is buffered at pH =3 with exactly sufficient Pb(NO_(3))_(2) to not precipitate Pbc. K_(a)of H_(2) S = 10^(-23),K_(sp) of Pbs = 10^(-28) . The concentration of Pb^(+2) in the solution is 10^(-x) . What is x ?

Answer» <html><body><p><br/></p>Solution :` H_2S hArr <a href="https://interviewquestions.tuteehub.com/tag/2h-300377" style="font-weight:bold;" target="_blank" title="Click to know more about 2H">2H</a>^(+)+S^(_2) rArr K_a =([H^(+)]^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>) [S^(-2)])/( [H_2S]) ` <br/> ` <a href="https://interviewquestions.tuteehub.com/tag/ph-1145128" style="font-weight:bold;" target="_blank" title="Click to know more about PH">PH</a> =3 rArr [H^(+) ] =10 ^(_3) M` <br/> ` 10 ^(-23) =( 10 ^(-6) [S^(-2)])/( 10^(-2) )rArr [S^(-2) ] =10 ^(-19)M`<br/> ` PbShArr Pb^(+2)+S^(-2)` <br/>` K_(sp)=[pb^(+2) ][S^(-2) ] rArr 10 ^(-28)=[pb^(+2) ] 10 ^(-19) `<br/> ` [pb^(+2) ] =10 ^(-<a href="https://interviewquestions.tuteehub.com/tag/9-340408" style="font-weight:bold;" target="_blank" title="Click to know more about 9">9</a>) M`</body></html>
Discussion
49477.

A sample was weighted using two different balances. The results were (i) 3.929 g and (ii) 4.0 g. How would the weight of the sample be reported?

Answer» <html><body><p>(a)` 3.929 g`<br/>(<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>)` 3 g`<br/>(<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a> )` 3.9 g`<br/>(d)`3.93 g`</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> :Round off the digit at 2nd position of <a href="https://interviewquestions.tuteehub.com/tag/decimal-435752" style="font-weight:bold;" target="_blank" title="Click to know more about DECIMAL">DECIMAL</a> 3.929=3.93.</body></html>
Discussion
49478.

A sample supposed to be pure CaCO_(3) is used to standardise a solution of HCl. The substance really was a mixture of MgCO_(3) and BaCO_(3), butthe standardisation of HCl was accurate. Find the percentage of BaCO_(3) and MgCO_(3) in mixture.

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> :`BaCO_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)=27.89%, MgCO_(3)=72.11%`;</body></html>
Discussion
49479.

A sample of zinc oxide contains 80.25% zinc. Calculate the equivalent mass of zinc.

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> :32.5</body></html>
Discussion
49480.

A sample of water has its hardness due to only CaSO_(4). When this water is passed through on anion exchange resin, SO_(4)^(2-) ions are replaced by OH^(-). A 25.0 mL sample of water so treated requires 21.58 mL of 10^(-3) MH_(2)SO_(4) for its titration. What is the hardness of water expressed in terms of CaCO_(3) in ppm? Assume density of water 1.0g//mL.

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> :`<a href="https://interviewquestions.tuteehub.com/tag/86-339622" style="font-weight:bold;" target="_blank" title="Click to know more about 86">86</a> "<a href="https://interviewquestions.tuteehub.com/tag/ppm-1162221" style="font-weight:bold;" target="_blank" title="Click to know more about PPM">PPM</a>"`;</body></html>
Discussion
49481.

A sample of water gas contains 42% by volume of carbon monoxide. If the total pressure is 760 mm. the partial pressure of carbon monoxide is

Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/380-310180" style="font-weight:bold;" target="_blank" title="Click to know more about 380">380</a> <a href="https://interviewquestions.tuteehub.com/tag/mm-1098795" style="font-weight:bold;" target="_blank" title="Click to know more about MM">MM</a> <br/>319.2 mm <br/>38 mm <br/>360 mm </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> :B</body></html>
Discussion
49482.

A sample of water contained 30ppm of MgSO_4 . Its hardness is

Answer» <html><body><p>25ppm<br/>20ppm <br/>30ppm<br/>150ppm</p>Answer :A</body></html>
Discussion
49483.

A sample of water contained 30 PPM of MgSO_(4) its hardness is x+20PPM What is 'x'

Answer» <html><body><p><br/></p>Solution :30 ppm of `MgSO_4 to 30 gm " in " 10^6` gm of `H_2O`<br/> Hardness in <a href="https://interviewquestions.tuteehub.com/tag/term-1241851" style="font-weight:bold;" target="_blank" title="Click to know more about TERM">TERM</a> of<br/> `CaCO_3 to 30 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 100/120 =2 <a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>`ppm</body></html>
Discussion
49484.

A sample of sucrose is found to contain 72.28 xx 10^21 atoms of carbon. Find the mass of the sample in grams.

Answer» <html><body><p></p>Solution :The <a href="https://interviewquestions.tuteehub.com/tag/molecular-562994" style="font-weight:bold;" target="_blank" title="Click to know more about MOLECULAR">MOLECULAR</a> formula of sucrose is `C_12H_22O_11`. From this formula, it is evident that one mole of sucrose <a href="https://interviewquestions.tuteehub.com/tag/contains-11473" style="font-weight:bold;" target="_blank" title="Click to know more about CONTAINS">CONTAINS</a> 12 gram atoms of carbon. The molecular mass of sucrose is`(12 xx 12.01) + (22 xx 1.008) + (11 xx 16.0) = 342.296` amu. <br/> Since, 1 gram atom of carbon contains `6.022 xx 10^(<a href="https://interviewquestions.tuteehub.com/tag/23-294845" style="font-weight:bold;" target="_blank" title="Click to know more about 23">23</a>)`atoms, the number of gram atoms <a href="https://interviewquestions.tuteehub.com/tag/corresponding-935567" style="font-weight:bold;" target="_blank" title="Click to know more about CORRESPONDING">CORRESPONDING</a> to <br/> `72.28 xx 10^(<a href="https://interviewquestions.tuteehub.com/tag/21-293276" style="font-weight:bold;" target="_blank" title="Click to know more about 21">21</a>)` atoms `=(72.28 xx 10^(21))/(6.022 xx 10^(23)) = 0.1200` <br/> `therefore` The gram moles of sucrose which contain 0.12 gram atoms of carbon `=1/12 xx 0.1200 = 0.0100` <br/> `therefore` The mass of 0.0100 moles of sucrose `=0.0100 xx 342.296 = 3.423 g`</body></html>
Discussion
49485.

A sample of sparingly soluble PbI_(2)(s) containing radioactive I-133 is added to 0.10M KI(aq) and stirred overnight. Observations about this system include which of the following? (P) The radioactivity of the liquid phase increases significantly. (Q) The concentration of the I^(-) ion in solution increases significantly.

Answer» <html><body><p>&lt;<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>&gt;P only<br/>R only<br/>Both P and Q<br/>Neither P nor Q</p>Answer :A</body></html>
Discussion
49486.

A sample of sour milk was found to be 0.1 M solution of lactic acid CH_(3) CH(OH)CO OH. What is the pH of the sample of milk ? K_(a) for lactic acid at 25^(@)C is 1.37xx10^(-4).

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> :2.43</body></html>
Discussion
49487.

A sample of solid KClO_(3) (potassium chlorate) was heated in a test tube to obtain O_(2) according to the reaction 2KClO_(3) rarr 2KCl + 3O_(2) The oxygen gas was collected by downward displacement of water at 295K. The total pressure of the mixture is 772 mm of Hg. The vapour pressure of water is 26.7 mm of Hg at 300K. What is the partial pressure of the oxygen gas ?

Answer» <html><body><p></p>Solution :`2KClO_(3(s)) <a href="https://interviewquestions.tuteehub.com/tag/rarr-1175461" style="font-weight:bold;" target="_blank" title="Click to know more about RARR">RARR</a> 2KCl_((s)) + 3O_(2 (g))` <br/> `P_("<a href="https://interviewquestions.tuteehub.com/tag/total-711110" style="font-weight:bold;" target="_blank" title="Click to know more about TOTAL">TOTAL</a>")=772mm <a href="https://interviewquestions.tuteehub.com/tag/hg-485049" style="font-weight:bold;" target="_blank" title="Click to know more about HG">HG</a>` <br/> `P_(H_(2)O) = 26.7 mm Hg` <br/> `P_("total") = P_(O_(2)) + P_(H_(2)O)` <br/> `:. P_(O_(2)) = P_("total") - P_(H_(2)O)` <br/> `P_(1) = 26.7 mm Hg T_(2) = 298K` <br/> `T_(1) = 300K, P_(2) = ?, (P_(1))/(T_(1)) = (P_(2))/(T_(2))` <br/> `P_(2) = ((P_(1))/(T_(1))) T_(2) = (26.7mm Hg)/(300 <a href="https://interviewquestions.tuteehub.com/tag/cancel-406753" style="font-weight:bold;" target="_blank" title="Click to know more about CANCEL">CANCEL</a>(K)) xx 295 cancel(K)` <br/> `P_(2) = 26.26 mm Hg` <br/> `:. P_(O_(2)) = 772-26.26` <br/> =745.74 mm Hg</body></html>
Discussion
49488.

A sample of sodium carbonate contains impurity of sodium sulphate. 1.25 g of this sample are dissolved in water and volume made up to 250 mL. 25 mL of this solution neutralise 20 mL of N/10sulphuric acid. Calculate the percentage of sodium carbonate in the sample.

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> :84.9</body></html>
Discussion
49489.

A sample of radioactive element undergoes 90% decomposition in 336 minutes. Its t_(0.5) in minutes is

Answer» <html><body><p>(In 2/l In 10) `xx <a href="https://interviewquestions.tuteehub.com/tag/366-1859606" style="font-weight:bold;" target="_blank" title="Click to know more about 366">366</a>`<br/><a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>/366<br/>(In 2/1 In 90) `xx 366`<br/>183</p>Solution :`t_(0.5)=(1)/(<a href="https://interviewquestions.tuteehub.com/tag/lambda-539003" style="font-weight:bold;" target="_blank" title="Click to know more about LAMBDA">LAMBDA</a>)"In" 2, t_(0.9)=(1)/(lambda)"In" 10` <br/> `therefore (t_(0.5))/(t_(0.9))=(ln 2)/(ln 10)` <br/> or `t_(0.5) = ("In"//2//"In" 10)xxt_(0.9)`</body></html>
Discussion
49490.

A sample of pynolusite (MnO_(2)) weights 0.5 gm. To this solutioin 0.594 gm As_(2)O_(3) and a dilute acid are added. After the reaction has stopped As^(+3) is AS_(2)O_(3) is titrated with 45 mlof M/50 KMn_(4) solution. Calculate the percentage of MnO_(2) in pyrolusite.

Answer» <html><body><p>`65.25%`<br/>`68%`<br/>`67.<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/>`66.6%`</p>Solution :`MnO_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)+As_(2)O_(3)rarrMn^(+2)+AsO_(4)^(-3)` <br/> For excess of `As_(2)O_(3)`, we have <br/> `As_(2)O_(3)+MnO_(4)^(-)rarrMn^(+2)+AsO_(4)^(-3)` <br/> Npw `As_(2)O_(3)` getting oxidized to `AsO_(4)^(-3)`, the half reaction is <br/> `As_(2)O_(3)+5H_(2)Orarr2As+10H^(+)+4e^(-)` <br/> for <a href="https://interviewquestions.tuteehub.com/tag/1mol-283039" style="font-weight:bold;" target="_blank" title="Click to know more about 1MOL">1MOL</a> of `As_(2)O_(3)`, No. of `e^(-)s=4`, <br/> So, eq. wt of `Al_(2)O_(3)=198//4` <br/> M.eq of `As_(2)O_(3)=(0.594)/(198//4)xx1000=12` <br/> Me eq of excess of `As_(2)O_(3)` = M.eq of `KMnO_(4)` <br/> `=45=((1)/(50)xx5)=4.5` <br/> The other half reaction is <br/> `MnO_(4)^(-)+5e^(-)+8H^(+)rarrMn^(2)+4H_(2)O` <br/> M.eq of `As_(2)O_(3)` used for `MnO_(2)=12-4.5=7.5` <br/> `therefore` wt implies `(W)/(87//2)xx1000=7.5impliesW=0.326gm` <br/> `%MnO_(2)=(0.326)/(0.5)xx100=65.25%`</body></html>
Discussion
49491.

A sample of pure PCl_(5) was introduced into an evacuted vessel at 473 K. After equilibrium ws attained , concentration of Pcl_(5) was found to be 0*5 xx 10^(-1)"mol" L^(-1). If value of K_(c)" is "8*3 xx 10^(-3), what are the concentrations of PCl_(3) and Cl_(2) at equilibrium ?

Answer» <html><body><p></p>Solution :` {:(,PCl_(<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>)(g),<a href="https://interviewquestions.tuteehub.com/tag/harr-2692945" style="font-weight:bold;" target="_blank" title="Click to know more about HARR">HARR</a>,PCl_(3)(g),+,Cl_(2)(g),), ("At eqm.", <a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>*5 xx 10^(-<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)"mol"L^(-1),,x"mol"L^(-1),,x"mol"L^(-1),("suppose")):}` <br/> ` :. K_(c) (x^(2))/(0*5 xx 10^(-1)) = 8*3 xx10^(-3) ("Given")` <br/> or `x^(2) = (8*3 xx 10^(-3))(0*5 xx 10^(-1)) = 4* <a href="https://interviewquestions.tuteehub.com/tag/15-274069" style="font-weight:bold;" target="_blank" title="Click to know more about 15">15</a> xx 10^(-4)`<br/>or ` x = sqrt(4*15 xx 10^(-4))= 2*04 xx 10^(-2)"M" = 0 *02 "M"`<br/> Hence, `[PCl_(3)]_(eq) = [Cl_(2)]_(eq) = 0*02 " M"`</body></html>
Discussion
49492.

A sample of pure PCl_5 was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl_5 was found to be 0.5 xx 10^(-1) "mol L"^(-1).If value of K_c is 8.3 xx 10^(-3), what are the concentrations of PCl_3and Cl_2 at equilibrium ? PCl_(5(g)) hArr PCl_(3(g)) +Cl_(2(g))

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> :At equilibrium `PCl_5(0.5xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>))=0.05 "mol L"^(-1)` <br/> Product `[Cl_2]`= Product `[PCl_3]`= x M so, <br/> `{:("<a href="https://interviewquestions.tuteehub.com/tag/balance-891682" style="font-weight:bold;" target="_blank" title="Click to know more about BALANCE">BALANCE</a> reaction:",PCl_(5(g)) hArr , PCl_(3(g)) + ,Cl_(2(g))),("Equilibrium <a href="https://interviewquestions.tuteehub.com/tag/concentration-20558" style="font-weight:bold;" target="_blank" title="Click to know more about CONCENTRATION">CONCENTRATION</a> :",0.5xx10^(-1),x ,x):}``K_c=([PCl_3][Cl_2])/([PCl_5])` <br/> `therefore 8.3xx10^(-3) =((x)(x))/(0.5xx10^(-1))` <br/> `therefore x^2=8.3xx10^(-3)xx0.5xx10^(-1)= 4.15xx10^(-4)` <br/> `therefore x=sqrt(4.15xx10^(-4))=2.037xx10^(-2)`=0.02037 M <br/> `therefore x approx 0.02 M = [PCl_3]=[Cl_2]`</body></html>
Discussion
49493.

A sample of pure carbon dioxide, irrespective of its source contains 27.27% carbon and 72.73% oxygen. The data support

Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/law-184" style="font-weight:bold;" target="_blank" title="Click to know more about LAW">LAW</a> of constant composition <br/>Law of conservtion of <a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> <br/>Law of reciprocal <a href="https://interviewquestions.tuteehub.com/tag/proportions-1170335" style="font-weight:bold;" target="_blank" title="Click to know more about PROPORTIONS">PROPORTIONS</a> <br/>Law of <a href="https://interviewquestions.tuteehub.com/tag/multiple-1105557" style="font-weight:bold;" target="_blank" title="Click to know more about MULTIPLE">MULTIPLE</a> proportions </p>Answer :A</body></html>
Discussion
49494.

A sample of protein was analysed for metal content and analysis revealed that it contained magnesium and titanium in equal amounts, by mass. If these are the only metallic species present in the protein and it contains 0.008% metal by mass, the minimum possible molar mass of the protein is : [Mg = 24, Ti = 48]

Answer» <html><body><p>`1.2 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> <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/22-294057" style="font-weight:bold;" target="_blank" title="Click to know more about 22">22</a>)`<br/>`1.2 xx 10^(<a href="https://interviewquestions.tuteehub.com/tag/25-296426" style="font-weight:bold;" target="_blank" title="Click to know more about 25">25</a>)`<br/>`7.2 xx 10^(21)`<br/>`1.08 xx 10^(22)`</p>Answer :D</body></html>
Discussion
49495.

A sample of pure Ca metal weighing 1.35 grams was quantitatively converted to 1.88 grams of pure Cao. What is the atomic weight of Ca? (O = 16)

Answer» <html><body><p></p>Solution :From the formula of Cao,<br/> we know, moles of <a href="https://interviewquestions.tuteehub.com/tag/ca-375" style="font-weight:bold;" target="_blank" title="Click to know more about CA">CA</a> = moles of O <br/> `("weight of Ca")/("atomic wt. of Ca")= ("weight of O")/(" atomic wt. of O")` <br/>`(1.35)/(at. wt. of Ca.) = ( 1.88 - 1.35)/(<a href="https://interviewquestions.tuteehub.com/tag/16-276476" style="font-weight:bold;" target="_blank" title="Click to know more about 16">16</a>)` <br/>Atomic weight of Ca = 40-75.</body></html>
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49496.

A sample of potato starch was ground in a ball mill to give a starchlike molecule of lower molecular weight. The product analysed 0.086% phosphorus. If each molecule is assumed to contain one atom of phosphorus, what is the molecular weight of the material?

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> :`3.6 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> <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/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)` <a href="https://interviewquestions.tuteehub.com/tag/amu-25369" style="font-weight:bold;" target="_blank" title="Click to know more about AMU">AMU</a></body></html>
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49497.

A sample of pond water contains 40mg of organic matter requires 32mg of dissolved oxygen. If pond water contains 100mg of organic matter per two litres, BOD value of the water sample is.

Answer» <html><body><p><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/ppm-1162221" style="font-weight:bold;" target="_blank" title="Click to know more about PPM">PPM</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/30-304807" style="font-weight:bold;" target="_blank" title="Click to know more about 30">30</a> ppm<br/>20 ppm<br/>40 ppm</p>Answer :D</body></html>
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49498.

A sample of pond water containing 20mg of organic matter requires 16mg of dissolved oxygen. (Pond water contains IOmg of organic matter per 2 litres). It's BOD is

Answer» <html><body><p>4000 <a href="https://interviewquestions.tuteehub.com/tag/ppm-1162221" style="font-weight:bold;" target="_blank" title="Click to know more about PPM">PPM</a><br/>400 ppm <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> ppm<br/>40 ppm</p>Solution :`O.D = (Wt of O_2)/("Wt. of <a href="https://interviewquestions.tuteehub.com/tag/sample-1194587" style="font-weight:bold;" target="_blank" title="Click to know more about SAMPLE">SAMPLE</a> of water") <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 10^6`</body></html>
Discussion
49499.

A sample of pond water containing 20mg of organic matter requires 16 mg of dissolved oxygen. (Pond water contains 10mg of organic matter per 2 litres). It's BOD is

Answer» <html><body><p>4000 ppm <br/>400 ppm <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> ppm <br/><a href="https://interviewquestions.tuteehub.com/tag/40-314386" style="font-weight:bold;" target="_blank" title="Click to know more about 40">40</a> ppm </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/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html>
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49500.

A sample of pithchbelen is foundto contain 50% uranium and 2.425% lead. Of this lead only 93% was Pb^(206) isotope. If the disintegration constant is 1.52xx10^(-10)yr how old could be the pithblende deposits?

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> :`3.3xx10^(<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a>)` <a href="https://interviewquestions.tuteehub.com/tag/year-1464599" style="font-weight:bold;" target="_blank" title="Click to know more about YEAR">YEAR</a></body></html>
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