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.
| 50601. |
A certain compound has the molecular formula X_(4)O_(6) having 57.2%X. Thus, |
| Answer» <html><body><p>atomic mass of X is 32<br/>X <a href="https://interviewquestions.tuteehub.com/tag/may-557248" style="font-weight:bold;" target="_blank" title="Click to know more about MAY">MAY</a> contain six <a href="https://interviewquestions.tuteehub.com/tag/valence-1442169" style="font-weight:bold;" target="_blank" title="Click to know more about VALENCE">VALENCE</a> electrons <br/>X is an electropositive metal <br/>X can be a non-metal</p>Solution :`57.2=(4x)/(M)xx100,43.8=(<a href="https://interviewquestions.tuteehub.com/tag/96-342541" style="font-weight:bold;" target="_blank" title="Click to know more about 96">96</a>)/(M)xx100impliesx=32`</body></html> | |
| 50602. |
A certain buffer solution contains equal concentration of X^(-) " and " HX.The K_(b) " for " X^(-) " is " 10^(-10) .Find the pH of the buffer . |
| Answer» <html><body><p></p>Solution :`k_(a) . k_(b)^(-<a href="https://interviewquestions.tuteehub.com/tag/14-272882" style="font-weight:bold;" target="_blank" title="Click to know more about 14">14</a>) "" :. K_(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>^(-14))/(10^(-10)) = 10^(-4)` <br/> `pH = pka + <a href="https://interviewquestions.tuteehub.com/tag/log-543719" style="font-weight:bold;" target="_blank" title="Click to know more about LOG">LOG</a>.([x^(-)])/([<a href="https://interviewquestions.tuteehub.com/tag/hx-492904" style="font-weight:bold;" target="_blank" title="Click to know more about HX">HX</a>]) ` <br/> ` :. pH = 4 + log. 1/1 = 4 "" :. pH = 4`</body></html> | |
| 50603. |
A certain buffer solution contains equal concentration of X^(-) and HX. K_(b), for X^(-) is 10^(-10). Find the pH of buffer. |
| 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>><br/></p>Solution :` K_b=<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a> ^(-10 ), Ka = 10 ^(-4)[X^(-) ]=[<a href="https://interviewquestions.tuteehub.com/tag/hx-492904" style="font-weight:bold;" target="_blank" title="Click to know more about HX">HX</a>]` <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>= P^(Ka) +<a href="https://interviewquestions.tuteehub.com/tag/log-543719" style="font-weight:bold;" target="_blank" title="Click to know more about LOG">LOG</a> ""([X^(-)])/([HX])=-log 10 ^(-4)=4`</body></html> | |
| 50604. |
(A): Cellulose is used as sizing agent in place of starch in textile industry. (R) : By changing process and raw-materials polluted waste is reduced. |
| 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 false <br/>(A) is false but (R) is true </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> :Both (A) and (R) are true and (R) is not the correct explanation of (A)</body></html> | |
| 50605. |
A cell is prepared by dipping copper rod in 1M copper suphate solutoin and zinc rod in 1M ZnSO_(4) solution The standard reduction potential s of copperand zinc are + 0.34 and -0.76 V respectively (i) what is the cell reaction ? (ii) what will be the standsard electromotive force (EMF) of the cell ? (iii) which electrode will be positive ? (iv) How will the cell be represented ? |
| Answer» <html><body><p><br/></p>Answer :(i) `Zn(s)+CuSO_(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)(aq)rarrZnSO_(4)(aq)+Cu(s);(ii)E_("cell")^(@)=1.1V`;(<a href="https://interviewquestions.tuteehub.com/tag/iii-497983" style="font-weight:bold;" target="_blank" title="Click to know more about III">III</a>)Cuelectrode (<a href="https://interviewquestions.tuteehub.com/tag/iv-501699" style="font-weight:bold;" target="_blank" title="Click to know more about IV">IV</a>)`Zn|ZnSO_(4)(1M)|CuSO_(4)(1M)|Cu`</body></html> | |
| 50606. |
A cell is prepared by dipping a chromium rod in 1 MCr_(2)(sO_(4))_(3) solution and an iron rod in 1 M FeSO_(4) solution The standard reduction potentials of chromium and iron electrodes ae -0.75 V and -0.45 V respecitvely (a) what will be the cell reactoition ? (b) what will be the standard EMF of hte cell ? (C ) which electrode will act as anode ? (d) which electrode will act as cathods ? |
| 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> :The two half cel reductio equation are : <br/> `2Fe^(2+)(aq) e^(-) rarr Fe(s),E^(@)=-0.45V` <br/> `Cr^(3+)(aq)+3 ^(-) rarr Cr(s),E^(@)=-0.75V` <br/> Since `Cr^(3+)//Cr` electrode has lower electrode <a href="https://interviewquestions.tuteehub.com/tag/potential-1161228" style="font-weight:bold;" target="_blank" title="Click to know more about POTENTIAL">POTENTIAL</a> therfore it acts as the anode while `Fe^(2+)//Fe` electrode with higher electrode potential acts as the cathode <br/> to equalise the number of electrons <a href="https://interviewquestions.tuteehub.com/tag/multiply-1106275" style="font-weight:bold;" target="_blank" title="Click to know more about MULTIPLY">MULTIPLY</a> Eq (i) by 3 and Eq (<a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a>) by 2 but do not multiply their `E^(@)` values <a href="https://interviewquestions.tuteehub.com/tag/thus-2307358" style="font-weight:bold;" target="_blank" title="Click to know more about THUS">THUS</a> <br/> To obtain equation for the cell reaction subtract Eq (iv) form Eq (iii) we have <br/> `2 Cr (s) +3Fe^(2+)(aq) rarr 2cr^(3+) (aq)+3 Fe(s) ,E_(cell)^(@) =-0.45 -(-0.75V)=+0.30 V` <br/> Thus the EmF of the cell =-+0.30 V</body></html> | |
| 50607. |
A cell is prepared by dipping a chromium rod is 1 M Cr_2(SO_4)_3 solution and an iron and in 1M FeSO_4 solutions. The standard potentials of chromium and iron electrodes are -0.75 V and 0.45 respectively. (a) What will be the cell reaction? (b) What will the standard EMF of the cell? (c) Which electrode will act an anode? (d) WHich electrode will act as cathode? |
| Answer» <html><body><p></p>Solution :The two half cell <a href="https://interviewquestions.tuteehub.com/tag/reduction-621019" style="font-weight:bold;" target="_blank" title="Click to know more about REDUCTION">REDUCTION</a> equations are: <br/> `Fe^(2+)(aq)+2e^(-) to Fe(s),E^@=-0.45V`....(i) <br/> `<a href="https://interviewquestions.tuteehub.com/tag/cr-427229" style="font-weight:bold;" target="_blank" title="Click to know more about CR">CR</a>^(3+)(aq)+3e^(-)toCr(s),E^@=-0.75 V`.....(ii) <br/>Since `Cr^(3+)//Cr` electrode has a power reduction <a href="https://interviewquestions.tuteehub.com/tag/potential-1161228" style="font-weight:bold;" target="_blank" title="Click to know more about POTENTIAL">POTENTIAL</a> it acts as the anode, while `Fe^(2+)//Fe` electrode with a higher electrode potential acts as the cathode. <br/> To equilize the number of electrons, multiply Eq (i) by 3 and Eq. (ii) by 2. But do not multiply their `E^@` values . Thus, <br/> `3Fe^(2+)(aq)+6e^(-)toFe(s),E^@=-0.45V`.....(<a href="https://interviewquestions.tuteehub.com/tag/iii-497983" style="font-weight:bold;" target="_blank" title="Click to know more about III">III</a>) <br/> `2Cr^(3r)(aq)+6e^(-)to2Cr(s),E^@=-0.75V`.....(iv)<br/> To obtain equations for the cell reactions, <a href="https://interviewquestions.tuteehub.com/tag/subtract-1231765" style="font-weight:bold;" target="_blank" title="Click to know more about SUBTRACT">SUBTRACT</a> Eq. (iv) from Eq. (iii) , we have, <br/> `2Cr(s)+3Fe^(2+)(aq)to2Cr^(3+)(aq)+3Fe(s)` <br/> `E_(cell)^o=-0.45-(-0.075V)=+0.30V` Thus, the EMF of the cell =+0.30 V</body></html> | |
| 50608. |
A cell is constructed using Cu^(2+)//Cu and I^(3+)//AIelectrode what is the net cell reaction |
| 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> :From we know that `E^(@)` of `AI^(3+)//AI(-1.66 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 `Cu^(2+)//Cu(+0.34 V)` <a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a> oxidation occurs at the `AI^(3+)//AI` electrode and reducation occurs at the `Cu^(2+) //Cu` electrode consequently the net cell reactin is <br/> `2 AI(s)+3 Cu^(2+)(aq)<a href="https://interviewquestions.tuteehub.com/tag/rarr-1175461" style="font-weight:bold;" target="_blank" title="Click to know more about RARR">RARR</a> 2 AI^(3+)(aq)+3Cu(s)`</body></html> | |
| 50609. |
A catalyst will increase the rate of a chemical reaction by lowering the ………. |
| 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/activation-848050" style="font-weight:bold;" target="_blank" title="Click to know more about ACTIVATION">ACTIVATION</a> <a href="https://interviewquestions.tuteehub.com/tag/energy-15288" style="font-weight:bold;" target="_blank" title="Click to know more about ENERGY">ENERGY</a></body></html> | |
| 50610. |
A catalyst remains unchanged at the end of the reaction regarding |
| Answer» <html><body><p>Mass<br/>Pysical <a href="https://interviewquestions.tuteehub.com/tag/state-21805" style="font-weight:bold;" target="_blank" title="Click to know more about STATE">STATE</a><br/>Physical state and <a href="https://interviewquestions.tuteehub.com/tag/chemical-914796" style="font-weight:bold;" target="_blank" title="Click to know more about CHEMICAL">CHEMICAL</a> composition<br/>Mass and chemical composition</p>Answer :D</body></html> | |
| 50611. |
A catalyst increases the speed of an equilibrium reaction and hence the amount of products formed Increases. |
| 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 <a href="https://interviewquestions.tuteehub.com/tag/catalyst-21033" style="font-weight:bold;" target="_blank" title="Click to know more about CATALYST">CATALYST</a> increases the speed of the forward as <a href="https://interviewquestions.tuteehub.com/tag/well-734398" style="font-weight:bold;" target="_blank" title="Click to know more about WELL">WELL</a> as backward reactions to the same <a href="https://interviewquestions.tuteehub.com/tag/extent-981159" style="font-weight:bold;" target="_blank" title="Click to know more about EXTENT">EXTENT</a>.</body></html> | |
| 50612. |
A catalyst is a substance which:- |
| Answer» <html><body><p>Alters the <a href="https://interviewquestions.tuteehub.com/tag/equilibrium-974342" style="font-weight:bold;" target="_blank" title="Click to know more about EQUILIBRIUM">EQUILIBRIUM</a> in a <a href="https://interviewquestions.tuteehub.com/tag/reaction-22747" style="font-weight:bold;" target="_blank" title="Click to know more about REACTION">REACTION</a><br/>Is always in the same phase as the reactants<br/>Participates in the reaction and provides easier <a href="https://interviewquestions.tuteehub.com/tag/pathway-1148972" style="font-weight:bold;" target="_blank" title="Click to know more about PATHWAY">PATHWAY</a> for the same<br/>Does not <a href="https://interviewquestions.tuteehub.com/tag/participate-590164" style="font-weight:bold;" target="_blank" title="Click to know more about PARTICIPATE">PARTICIPATE</a> in the reaction but speeds it up.</p>Solution :A catalyst does not take part I the reaction but can speed it up. It can be <a href="https://interviewquestions.tuteehub.com/tag/recovered-2246317" style="font-weight:bold;" target="_blank" title="Click to know more about RECOVERED">RECOVERED</a> after the reaction.</body></html> | |
| 50613. |
A catalyst in the finely divided state is more efficient because in this state |
| Answer» <html><body><p>it has <a href="https://interviewquestions.tuteehub.com/tag/larger-1067345" style="font-weight:bold;" target="_blank" title="Click to know more about LARGER">LARGER</a> <a href="https://interviewquestions.tuteehub.com/tag/activation-848050" style="font-weight:bold;" target="_blank" title="Click to know more about ACTIVATION">ACTIVATION</a> energy<br/>it can react with one of the <a href="https://interviewquestions.tuteehub.com/tag/reactant-1178091" style="font-weight:bold;" target="_blank" title="Click to know more about REACTANT">REACTANT</a> more efficiently<br/>it has large surface <a href="https://interviewquestions.tuteehub.com/tag/area-13372" style="font-weight:bold;" target="_blank" title="Click to know more about AREA">AREA</a><br/>All the above</p>Solution :More the surface area , more is the number of <a href="https://interviewquestions.tuteehub.com/tag/active-367234" style="font-weight:bold;" target="_blank" title="Click to know more about ACTIVE">ACTIVE</a> centres and hence more is the efficiency of the catalyst.</body></html> | |
| 50614. |
A catalyst can affect reversible reaction by:- |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/changing-2508325" style="font-weight:bold;" target="_blank" title="Click to know more about CHANGING">CHANGING</a> <a href="https://interviewquestions.tuteehub.com/tag/equilibrium-974342" style="font-weight:bold;" target="_blank" title="Click to know more about EQUILIBRIUM">EQUILIBRIUM</a><br/>Slowing forward reaction<br/>Attaining equilibrium in both direction<br/>None of these</p>Answer :C</body></html> | |
| 50615. |
A catalyst |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/alters-858771" style="font-weight:bold;" target="_blank" title="Click to know more about ALTERS">ALTERS</a> the equilibrium constant<br/>Increases the 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> of products<br/>helps establishing the equilibrium quickly<br/> Supplies energy to the reactants</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/catalyst-21033" style="font-weight:bold;" target="_blank" title="Click to know more about CATALYST">CATALYST</a> <a href="https://interviewquestions.tuteehub.com/tag/properties-11511" style="font-weight:bold;" target="_blank" title="Click to know more about PROPERTIES">PROPERTIES</a></body></html> | |
| 50616. |
(A): CaSO_(4) is called dead burnt gypsum because it does not set with water (R) : Mixture of 1 part of slaked lime, 3 parts of sand and water is known asmortar |
| Answer» <html><body><p>Both A and <a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a> are correct 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 and R are correct but R is not the correct explanation of A. <br/>A is True but R is False. <br/>R is False but A 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> | |
| 50617. |
(A) : Carbon forms a large number of compounds (R) : Carbon has high catenation power |
| Answer» <html><body><p>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>, R explains A <br/>A and R are true, R does not explain 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> | |
| 50618. |
(a ) Carbon dioxide is non -polar while water is polar. What conclusion do you draw about their structure from for this fact ? (b) Classify the following compounds into acidic basic and amphoteric oxides . AI_(2) O_(3),CI_(2)O_(7) . |
| 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 ) `Ooverset rarr =Coversetrarr = O` is <a href="https://interviewquestions.tuteehub.com/tag/linear-1074458" style="font-weight:bold;" target="_blank" title="Click to know more about LINEAR">LINEAR</a> bond moments are equal and opposite net dipole moment is zero. Water is a <a href="https://interviewquestions.tuteehub.com/tag/bent-395665" style="font-weight:bold;" target="_blank" title="Click to know more about BENT">BENT</a> molecule it has a net dipole moment . <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/SPH_BSR_CHE_QB_XI_C11_E03_017_S01.png" width="80%"/> <br/> ( b) `AI_(2)O_(3)` is <a href="https://interviewquestions.tuteehub.com/tag/amphoteric-374858" style="font-weight:bold;" target="_blank" title="Click to know more about AMPHOTERIC">AMPHOTERIC</a> where as `CI_(2) O_(7)` is an acidic <a href="https://interviewquestions.tuteehub.com/tag/oxide-1144484" style="font-weight:bold;" target="_blank" title="Click to know more about OXIDE">OXIDE</a>.</body></html> | |
| 50619. |
A carbon compound on analysis gave the following percentage composition, carbon 14.5%, hydrogen 1.8%, chlorine 64.46%, oxygen 19.24%. Calculate the empirical formula of the compound. |
| 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> :`C_(<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_(3)Cl_(3)O_(2)`</body></html> | |
| 50620. |
A carbon compound has many functional groups, then order of preference while naming it according to IUPAC nomenclature is |
| Answer» <html><body><p>`-<a href="https://interviewquestions.tuteehub.com/tag/cho-408392" style="font-weight:bold;" target="_blank" title="Click to know more about CHO">CHO</a> <a href="https://interviewquestions.tuteehub.com/tag/gt-1013864" style="font-weight:bold;" target="_blank" title="Click to know more about GT">GT</a> - COOH gt -<a href="https://interviewquestions.tuteehub.com/tag/oh-585115" style="font-weight:bold;" target="_blank" title="Click to know more about OH">OH</a> gt -NH_(2)`<br/>` -COOH gt -CHO gt -NH_(2) gt -OH`<br/>`-COOH gt - OH gt - NH_(2) gt -CHO`<br/>`-COOH gt -CHO gt -OH gt - NH_(2)`</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> | |
| 50621. |
A carbon compound contains 12.8% Carbon, 2.1% Hydrogen, 85.1% Bromine. The molecular weight of the compound is 187.9. Calculate the molecular formula. |
| Answer» <html><body><p></p>Solution :Step <a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>: Percentage composition of the <a href="https://interviewquestions.tuteehub.com/tag/elements-969381" style="font-weight:bold;" target="_blank" title="Click to know more about ELEMENTS">ELEMENTS</a> present in the <a href="https://interviewquestions.tuteehub.com/tag/compound-926863" style="font-weight:bold;" target="_blank" title="Click to know more about COMPOUND">COMPOUND</a>. <br/> `{:(C,H,Br),(12.8,2.1,85.1):}` <br/> Step 2: Dividing with the respective atomic <a href="https://interviewquestions.tuteehub.com/tag/weights-1451449" style="font-weight:bold;" target="_blank" title="Click to know more about WEIGHTS">WEIGHTS</a> of the elements. <br/> `{:((12.8)/(12),(2.1)/(1),(85.1)/(80)),(1.067,2.1,1.067):}` <br/> Step 3: Dividing by the smallest number to get simple atomic <a href="https://interviewquestions.tuteehub.com/tag/ratio-13379" style="font-weight:bold;" target="_blank" title="Click to know more about RATIO">RATIO</a>. <br/> `{:((1.067)/(1.067),(2.1)/(1.067),(1.067)/(1.067)),(1,2,1):}` <br/> The empirical formula is `CH_(2)Br` <br/> Empricial formula weight `12 + (2 xx 1) + 80 = 94` <br/> The molecular weight = 187.9 (given) <br/> `therefore n=(187.9)/(94)=2` <br/> `"The molecular formula" = "(empirical formula)"_(2)` <br/> `(CH_(2)Br)_(2)=C_(2)H_(4)Br_(2)`</body></html> | |
| 50622. |
A carbon compound containing only carbon and oxygen has an approximate molecular weight of 290. On analysis it is found to contain 50% by weight of each element. What is the molecular formula of the compound ? |
| Answer» <html><body><p><br/></p>Solution :Calculation of empirical formula : <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/NTN_HCS_ISC_CHE_XI_P1_C01_E09_020_S01.png" width="80%"/> <br/> `therefore` the empirical formula `=C_(4)O_(3)` <br/> Calculations of <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: <br/> Molecular <a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> = 290 (given) <br/> Empirical formula mass `=(4 xx 12.01 + (3 xx 16.0) = 96.04` <br/> `therefore <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a>=("Molecular mass")/("Empirical formula mass") = 290/(96.04) = 3.019 =3` <br/> `therefore` The molecular formula of the given compound <br/> `= 3 xx C_(4)O_(3) = C_(12)O_(<a href="https://interviewquestions.tuteehub.com/tag/9-340408" style="font-weight:bold;" target="_blank" title="Click to know more about 9">9</a>)`</body></html> | |
| 50623. |
(A) : Carbon-carbon bond energy values are in the order C_(2)H_(2) gt C_(2)H_(4) gt C_(2)H_(6) (R ): Bond energy increases with increase in the bond order |
| Answer» <html><body><p>A and R are true, R explains A<br/>A and R are true, R does not explain 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>Answer :A</body></html> | |
| 50624. |
A carbocation in which dispersal of charge does not take place |
| Answer» <html><body><p>`R_(3)C^(oplus)`<br/>`R_(3)<a href="https://interviewquestions.tuteehub.com/tag/ch-913588" style="font-weight:bold;" target="_blank" title="Click to know more about CH">CH</a>^(oplus)`<br/>`RCH_(2)^(oplus)`<br/>`""^(oplus)CH_(3)`</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/among-374776" style="font-weight:bold;" target="_blank" title="Click to know more about AMONG">AMONG</a> the given cabo cataions there is no <a href="https://interviewquestions.tuteehub.com/tag/dispersal-956024" style="font-weight:bold;" target="_blank" title="Click to know more about DISPERSAL">DISPERSAL</a> of charge in `CH_(3)^(+)` (or) No-hyper conjugation</body></html> | |
| 50625. |
(a) Calculate the energy needed to raise the temperature of10.0 g of iron from 25^(@) to 500^(@)C if specific heat capacity of iron is 0.45 J (^(@C) g^(-1). (b) What mass of gold of specific heat capacity 0.13 J ( .^(@)C)^(-1) g^(-1) can be heated through the same temperature difference when supplied with the same amount of energyas in (a) ? |
| 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) Energy needed `(<a href="https://interviewquestions.tuteehub.com/tag/q-609558" style="font-weight:bold;" target="_blank" title="Click to know more about Q">Q</a>) =m <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/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a> xx Delta T = 10.0xx 0.45xx ( 500 -25) J = 2137.5 J` <br/> (b) `q= m xx c xx Delta T` <br/>`2137.5 = m xx 0.13 xx ( 500 - 25) ` or ` m = 34.6 g`</body></html> | |
| 50626. |
(A): C Cl_(2) does not undergo hydrolysis where as SiCl_(4) is readily hydrolysed. (R) : Carbon has no d-orbitals in its valence shell, but silicon has vacant d-orbitals in its valence shell. |
| Answer» <html><body><p>a. Both A and <a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a> are true, and R is correct explanation of A <br/>b. Both A and R are true, and R is not correct explanation of A <br/>c. A is true, but R is false<br/>d. 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> | |
| 50627. |
Butyric acid contains only C, H and O. A 4.24 mg sample of butyric acid is completely burnt. It gives 8.45 mg of carbon dioxide and 3.46 mg of water. What is the mass percentage of each element in butyric acid? |
| Answer» <html><body><p><br/></p>Answer :( c) <a href="https://interviewquestions.tuteehub.com/tag/molecular-562994" style="font-weight:bold;" target="_blank" title="Click to know more about MOLECULAR">MOLECULAR</a> <a href="https://interviewquestions.tuteehub.com/tag/formula-464310" style="font-weight:bold;" target="_blank" title="Click to know more about FORMULA">FORMULA</a> `=C_(4)H_(<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a>)O_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)`</body></html> | |
| 50628. |
A burning strip of magnesium is introduced into a jar containing a gas. After sometime the walls of the container is coated with carbon. The gas in the container is |
| Answer» <html><body><p>`O_(<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/>`N_(2)`<br/>`CO_(2)`<br/>`H_(2)O`</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> :C</body></html> | |
| 50629. |
A burning magnesium ribbon will continue to burn in |
| Answer» <html><body><p>`CO_(<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 `SO_(2)`<br/>`N_(2),CO_(2)` and `SO_(2)`<br/>`N_(2)` and steam<br/>`N_(2)`, `CO_(2)`, `SO_(2)` and steam</p>Solution :A burning .Mg. ribbon will continue to burn in the <a href="https://interviewquestions.tuteehub.com/tag/atmospheres-887162" style="font-weight:bold;" target="_blank" title="Click to know more about ATMOSPHERES">ATMOSPHERES</a> of `N_(2)`, `CO_(2)`, `SO_(2)` steam and <a href="https://interviewquestions.tuteehub.com/tag/even-976335" style="font-weight:bold;" target="_blank" title="Click to know more about EVEN">EVEN</a> in `N_(2)O`.</body></html> | |
| 50630. |
A bulb of unknown volume 'V' Contains an ideal gas at 2 atm pressure. It was connected to another evacuated bulb of volume 0.5 litre through a stopcock. When the stopcock was opened the pressure in each bulb became 0.5 atm. Then V is |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/21-293276" style="font-weight:bold;" target="_blank" title="Click to know more about 21">21</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> c `<br/>`42c ,c`<br/>`105c c `<br/>`<a href="https://interviewquestions.tuteehub.com/tag/63c-1909288" style="font-weight:bold;" target="_blank" title="Click to know more about 63C">63C</a> c `</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> | |
| 50631. |
A bulb emits light of wavelength 4500 Å. The bulb is rated as 150 watt and 8% of the energy is emitted as light. How many photons are emitted by the bulb per second ? |
| Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/electrical-968148" style="font-weight:bold;" target="_blank" title="Click to know more about ELECTRICAL">ELECTRICAL</a> energy in joules = Power in watts `xx` Time in seconds <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> <a href="https://interviewquestions.tuteehub.com/tag/150-275254" style="font-weight:bold;" target="_blank" title="Click to know more about 150">150</a> watt = 150 joules of energy emitted per second <br/> `:.` Energy emitted as light `= (8)/(100) xx 150 = 12J` <br/> `E = n <a href="https://interviewquestions.tuteehub.com/tag/hv-1033771" style="font-weight:bold;" target="_blank" title="Click to know more about HV">HV</a> = nh (c)/(lamda) " " :. n = (E xx lamda)/(h xx c) = ((12J) xx (4500 xx 10^(-10) m))/((6.626 xx 10^(-34) Js) xx (3 xx 10^(8) ms^(-1))) = 2.717 xx 10^(19)`</body></html> | |
| 50632. |
A bulb emits light of wave length 4500Å. The bulb is rated as 150 watt and 8% of the energy is emitted as light. How many photons are emitted by the bulb per second ? |
| 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> :`n=27.2 <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/18-278913" style="font-weight:bold;" target="_blank" title="Click to know more about 18">18</a>)`</body></html> | |
| 50633. |
A bulb containing N_(2)O_(4) is colourless in ice. Its colour inboiling water is …………. while in water at 298 K, it is………… . |
| 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/reddish-1181269" style="font-weight:bold;" target="_blank" title="Click to know more about REDDISH">REDDISH</a> <a href="https://interviewquestions.tuteehub.com/tag/brown-904764" style="font-weight:bold;" target="_blank" title="Click to know more about BROWN">BROWN</a>, <a href="https://interviewquestions.tuteehub.com/tag/pale-590054" style="font-weight:bold;" target="_blank" title="Click to know more about PALE">PALE</a> brown</body></html> | |
| 50634. |
A buffer solution with pH 9 is to be prepared by mixing NH_(4)Cl that should be added to one litre of 1.0mNH_(4)OH kb1.8 xx 10^(-5) |
| 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> :`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>)<a href="https://interviewquestions.tuteehub.com/tag/cl-408888" style="font-weight:bold;" target="_blank" title="Click to know more about CL">CL</a> = 1.8 M`</body></html> | |
| 50635. |
A buffer solution with pH 9 is to be prepared by mixing NH_(4) Cl and NH_(4)OH. Calculate the number of moles of NH_(4)Cl that should be added to one litre of 1.0 M NH_(4)OH (K_(b)=1.8xx10^(-5)) |
| Answer» <html><body><p></p>Solution :For a <a href="https://interviewquestions.tuteehub.com/tag/basic-15343" style="font-weight:bold;" target="_blank" title="Click to know more about BASIC">BASIC</a> <a href="https://interviewquestions.tuteehub.com/tag/buffer-905159" style="font-weight:bold;" target="_blank" title="Click to know more about BUFFER">BUFFER</a>, `pOH = pK_(b)+log.(["<a href="https://interviewquestions.tuteehub.com/tag/salt-1193804" style="font-weight:bold;" target="_blank" title="Click to know more about SALT">SALT</a>"])/(["Base"])` <br/> Further, `pH + pOH = 14 ` so that `pOH = 14 - pH = 14 - 9 = 5` <br/> `pK_(b)=-log K_(b) = - log(1.8xx10^(-5))=4.7447` <br/> [Base] `=[NH_(4)OH]= 1 "<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)` <br/> `:. 5 = 4. 7447 + log. (["Salt"])/(1) or log ["Salt"] = 0.2553 or ["Salt"]= 1.8 "mol" L^(-1)`</body></html> | |
| 50636. |
A buffer solution of pH = 4.7 is prepared from CH_(3)COONa and CH_(3)COOH. Dissociation constant of acetic acid is 1.75xx10^(-5). Calculate the mole proportion of sodium acetate and acetic acid. |
| Answer» <html><body><p></p>Solution :`pH=pK_(a)+"log"(["<a href="https://interviewquestions.tuteehub.com/tag/salt-1193804" style="font-weight:bold;" target="_blank" title="Click to know more about SALT">SALT</a>"])/(["Acid"])rArrpH=-log_(10)K_(a)+"log"([CH_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)COONa])/([CH_(3)<a href="https://interviewquestions.tuteehub.com/tag/cooh-409857" style="font-weight:bold;" target="_blank" title="Click to know more about COOH">COOH</a>])` <br/> `4.7=-log_(10)(1.75xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>))+"log"([CH_(3)COONa])/([CH_(3)COOH])` <br/> `4.7=5-log1.75+"log"([CH_(3)COONa])/([CH_(3)COOH])=5-0.2430+"log"([CH_(3)COONa])/([CH_(3)COOH])` <br/> `"log"([CH_(3)COONa])/([CH_(3)COOH])="anti"log(-0.0557)="anti"log(-0.057+1-1)` <br/> `([CH_(3)COONa])/([CH_(3)COOH])=0.8770`.</body></html> | |
| 50637. |
A buffer solution of pH 8.3 is prepared from ammonium chloride and ammonium hydroxide. Dissociation constant of ammonium hydroxide is 1.8xx10^(-5). What is the mole proportion of ammonium chloride and ammonium hydroxide? |
| Answer» <html><body><p></p>Solution :pOH = 14 - pH = 14 - 8.3 = 5.7 <br/> `pH=pK_(a)+"log"(["<a href="https://interviewquestions.tuteehub.com/tag/salt-1193804" style="font-weight:bold;" target="_blank" title="Click to know more about SALT">SALT</a>"])/(["Base"])rArrpH=-log_(<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>)K_(a)+"log"(["Salt"])/(["Base"])` <br/> `5.7=-log_(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>.8xx10^(-5))+"log"([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>])/([NH_(4)OH])rArr5.7=5-log1.8+"log"([NH_(4)Cl])/([NH_(4)OH])` <br/> `5.7=5-0.2553+"log"([NH_(4)Cl])/([NH_(4)OH])rArr5.7=4.7447+"log"([NH_(4)Cl])/([NH_(4)OH])` <br/> `therefore([NH_(4)Cl])/([NH_(4)OH])=5.7-4.7447=0.9553` <br/> `rArr([NH_(4)Cl])/([NH_(4)OH])="anti"log(0.9553)=9.022""` The ratio is 9 : 1.</body></html> | |
| 50638. |
A buffer solution of acetic acid and sodium acetate is diluted 10 times. What is the effect on itspH ? |
| Answer» <html><body><p></p>Solution :No effect because`pH = pK_(a) + <a href="https://interviewquestions.tuteehub.com/tag/log-543719" style="font-weight:bold;" target="_blank" title="Click to know more about LOG">LOG</a>. (["Salt"])/(["<a href="https://interviewquestions.tuteehub.com/tag/acid-847491" style="font-weight:bold;" target="_blank" title="Click to know more about ACID">ACID</a>"])`. On dilution, ratio [Salt]/[Acid] remains same. Hence, pHdoes not change.</body></html> | |
| 50639. |
A buffer solution is prepared in which the concentration of NH_(3) is 0.30 M and the concentrationof NH_(4)^(+) is 0.20 M. If the equilibrium constant, K_(b) for NH_(3) equals 1.8xx10^(-5), what is the pH of this solution ? |
| 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>>8.73<br/>9.08<br/>9.43<br/>11.72</p>Solution :`p_(<a href="https://interviewquestions.tuteehub.com/tag/oh-585115" style="font-weight:bold;" target="_blank" title="Click to know more about OH">OH</a>)=pK_(<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>)+log.(["Salt"])/(["<a href="https://interviewquestions.tuteehub.com/tag/base-892693" style="font-weight:bold;" target="_blank" title="Click to know more about BASE">BASE</a>"])=4.74+log.(0.20)/(0.30)` <br/> `=4.74+(0.301-0.4.74)=4.74-0.176` <br/> `=4.56` <br/> `:. pH = 14 - 4.56 = 9.44`</body></html> | |
| 50640. |
A buffer solution is prepared by mixing CH_(3) COONa and CH_(3) COOH. The pK_(a) of acetic acid is 4.74 . To maintain the pH of the buffer solution as 6.04 , the concentration ratio of CH_(3) COONa and CH_(3) COOH to be maintained is |
| 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> : <a href="https://interviewquestions.tuteehub.com/tag/20-287209" style="font-weight:bold;" target="_blank" title="Click to know more about 20">20</a>`<br/>`1 : <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>`<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> : 1`<br/>`20 : 1`</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> | |
| 50641. |
A buffer solution is formeed by mixing 100 mL of 0.1 M CH_(3)COOH with 200 mL of 0.02 M CH_(3)COONa. If this buffer solution is made to 1.0 L by adding 700 mL of water, pH will change by a factor of |
| Answer» <html><body><p><br/></p>Solution :No change in pH of buffer <a href="https://interviewquestions.tuteehub.com/tag/sol-1216281" style="font-weight:bold;" target="_blank" title="Click to know more about SOL">SOL</a>. By <a href="https://interviewquestions.tuteehub.com/tag/dilution-439763" style="font-weight:bold;" target="_blank" title="Click to know more about DILUTION">DILUTION</a></body></html> | |
| 50642. |
A buffer solutionis prepared by mixing 50 mL of 0.2 Mof acetic acid with .x. mL of 0.2 M of NaOHsolutions. If pH of the resulting buffersolutions is 4. 7 then value of .x.is (Ka= 2xx10 ^(-5)) |
| Answer» <html><body><p>28 mL <br/> 22 m L <br/>24 mL <br/><a href="https://interviewquestions.tuteehub.com/tag/25-296426" style="font-weight:bold;" target="_blank" title="Click to know more about 25">25</a> mL </p>Solution :` pH =pKa +log "" ([S])/([A]), 4.7 =4.7 +log "" ([S])/([A]) ` <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]=[A]` <br/> ` x <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 0.2 = 50 xx 0.2 - x xx 0.2 ` <br/> `rArr x = ( 50 xx 0.2)/(2 xx 0.2 )= 25 mL `</body></html> | |
| 50643. |
A buffer solution contains a weak acid HA and A . When small quantity of NaOH is added, to keep p^(H)as constant. which of the following reaction takes place? |
| Answer» <html><body><p>` <a href="https://interviewquestions.tuteehub.com/tag/ha-479237" style="font-weight:bold;" target="_blank" title="Click to know more about HA">HA</a> to H^(+) A^(-) ` <br/>` H^(+) A^(-) toHA` <br/>` HA+ <a href="https://interviewquestions.tuteehub.com/tag/oh-585115" style="font-weight:bold;" target="_blank" title="Click to know more about OH">OH</a>^(-) to H_2O +A^(-) ` <br/>`A^(-) +H_2O to HA+ OH^(-) ` </p>Solution :NaOHadded is a base hence acid <a href="https://interviewquestions.tuteehub.com/tag/component-926634" style="font-weight:bold;" target="_blank" title="Click to know more about COMPONENT">COMPONENT</a> of the buffter . `HA+ OH^(-)to H_2O+A^(-) `</body></html> | |
| 50644. |
A buffer solution contains 0.40 mol of ammonium hydroxide and 0.50 mol of ammonium chloride to make a buffer solution of 1 L. Calculate the pH of the resulting buffer solution. Dissociation constant of ammonium hydroxide at 25^(@)C is 1.81xx10^(-5). |
| 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> :`pOH = pK_(b) + log. (["Salt"])/(["Base"])=pK_(b) + log. ([NH_(4)Cl])/([NH_(4)OH])` <br/>`[NH_(4)OH]=0.40` mol `L^(-1)` <br/> `[NH_(4)Cl]=0.50 ` mol `L^(-1)` <br/> `pK_(b) = - log K_(b) = - log (1.81 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/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>))=5-0.2577 = 4.7423` <br/> `:. pOH = 4.742 + log. (0.5)/(0.4)= 4.742+ log 1.25 = 4.742 + 0.0969 = 4.8389 ~= 4.839` <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> = 14-pOH = 14 - 4.839 = 9.161`.</body></html> | |
| 50645. |
A buffer solution contains 2 moles of ammonium hydroxide and 0.25 mole of ammonium chloride per dm^(3) of the solution. (K_(b) for ammonium hydroxide = 1.8xx10^(-5)). Calculate the pH of the buffer solution. |
| Answer» <html><body><p></p>Solution :It is a <a href="https://interviewquestions.tuteehub.com/tag/basic-15343" style="font-weight:bold;" target="_blank" title="Click to know more about BASIC">BASIC</a> <a href="https://interviewquestions.tuteehub.com/tag/buffer-905159" style="font-weight:bold;" target="_blank" title="Click to know more about BUFFER">BUFFER</a> <br/> `pOH=pK_(<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 href="https://interviewquestions.tuteehub.com/tag/log-543719" style="font-weight:bold;" target="_blank" title="Click to know more about LOG">LOG</a>"(["Salt"])/(["Base"])` <br/> = `4.74+"log"0.25/2=4.74+"log"0.25/2` <br/> = 4.74 - 0.90 = 3.64 <br/> pH = 14 - 3.64 <br/> = 10.36</body></html> | |
| 50646. |
A buffer solution contains 0.2mole of NH_4 OHand 0.2 mole of NH_4 Cl perlitre. The PK_bofNH_4 OHis 4.75 The pH of the buffer will be |
| Answer» <html><body><p>`4.75` <br/>` 5.75` <br/>`9.25`<br/>`2.25`</p>Solution :` pOH =pK_b +<a href="https://interviewquestions.tuteehub.com/tag/log-543719" style="font-weight:bold;" target="_blank" title="Click to know more about LOG">LOG</a> "" ([S])/([B])=4.75 +log ""(0.2)/(0.2) = 4.75` <br/> `rArr <a href="https://interviewquestions.tuteehub.com/tag/ph-1145128" style="font-weight:bold;" target="_blank" title="Click to know more about PH">PH</a> = 14 -4.75 =9.25`</body></html> | |
| 50647. |
A buffer solution contains 0.1 mole of sodium acetate in 1000 cm^(3) of 0.1 M acetic acid. To the above buffer solution, 0.1 M aceticacid. To the above buffer solution, 0.1 mole of sodium acetate is further added and dissolved. The pH of the resultingbuffer is equal to ........... |
| Answer» <html><body><p>`pK_(a) - log2`<br/>`pK_(a)`<br/>`pK_(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/>`pK_(a)+<a href="https://interviewquestions.tuteehub.com/tag/log-543719" style="font-weight:bold;" target="_blank" title="Click to know more about LOG">LOG</a> 2`</p>Solution :`pH = pK_(a) + log.(["Salt"])/(["Acid"])` <br/> [Salt]`=(0.1+0.1"mole")/(1000 <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a>) = (0.2"mole")/(<a href="https://interviewquestions.tuteehub.com/tag/1l-282810" style="font-weight:bold;" target="_blank" title="Click to know more about 1L">1L</a>) = 0.2M` <br/> [Acid]`=(0.1 "mole")/(1000mL) = (0.1"mole")/(1L)=0.1 M` <br/> `:. pH = pK_(a) + log 2`.</body></html> | |
| 50648. |
A buffer solution consists of 1 mole each of HA and its conjugate base A^(-), Addition of which of the following decreases the pH of buffer ? |
| Answer» <html><body><p>0.1 <a href="https://interviewquestions.tuteehub.com/tag/mole-1100299" style="font-weight:bold;" target="_blank" title="Click to know more about MOLE">MOLE</a> of <a href="https://interviewquestions.tuteehub.com/tag/ha-479237" style="font-weight:bold;" target="_blank" title="Click to know more about HA">HA</a> <br/>0.1 mole of `A^(-)`<br/>0.1 mole of `<a href="https://interviewquestions.tuteehub.com/tag/naoh-572531" style="font-weight:bold;" target="_blank" title="Click to know more about NAOH">NAOH</a>`<br/><a href="https://interviewquestions.tuteehub.com/tag/water-1449333" style="font-weight:bold;" target="_blank" title="Click to know more about WATER">WATER</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</body></html> | |
| 50649. |
A buffer solution can be prepared from a mixture 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> acetate and acetic acid in water<br/>sodium acetate and <a href="https://interviewquestions.tuteehub.com/tag/hydrochloric-493108" style="font-weight:bold;" target="_blank" title="Click to know more about HYDROCHLORIC">HYDROCHLORIC</a> acid in water<br/>ammonia and ammonium chloride in water<br/>ammonia and sodium hydroxide in water</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 buffer solution is a mixture of <a href="https://interviewquestions.tuteehub.com/tag/weak-729638" style="font-weight:bold;" target="_blank" title="Click to know more about WEAK">WEAK</a> acid/base and its salt with conjugate base/acid (<a href="https://interviewquestions.tuteehub.com/tag/strong-653928" style="font-weight:bold;" target="_blank" title="Click to know more about STRONG">STRONG</a> base /acid)</body></html> | |
| 50650. |
A buffer solution 0.04 M in Na_2HPO_4 and 0.02M in Na_3 PO_4 is prepared. The electrolytic oxidation of 1.0 milli -mole of the organic compound RNHOH is carried out in 100 mLof the buffer. The reaction is RNHOH +H_2O to RNO_2 +4H ^(+) + 4e ^(-) The approximatepH of solution after the oxidation is complete is:[Given : forH_3O PO_4, pK_(a_1)=7.20 , pK_(a_2) =12] |
| Answer» <html><body><p>` 6.90 ` <br/>` <a href="https://interviewquestions.tuteehub.com/tag/7-332378" style="font-weight:bold;" target="_blank" title="Click to know more about 7">7</a>.20 ` <br/>` 7. <a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>` <br/>None of these </p>Solution :` {:(PO_4^(3-)+,H^(+)to, HPO_4^(-2)),( <a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>.02M , 0.04 M , 0.04 M),( -, 0.02 , 0.06M), (- , 0.02M, -):}` <br/> ` {:( HPO_4^(-2) ,H^(+) to, H_2PO_4^(-)),( 0.06, 0.02 , 0),( 0.02 , - , 0.02M),( 0.04M, , ):}` <br/> ` pH =7.2 +<a href="https://interviewquestions.tuteehub.com/tag/log-543719" style="font-weight:bold;" target="_blank" title="Click to know more about LOG">LOG</a> ""(0.04)/(0.02 )=7.5`</body></html> | |