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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
An ideal solution was obtained by mixing methanol and ethanol. If the partial vapour pressure of methanol and ethanol are `2.619 kPa` and `4.556 kPa`, respectively, the composition of vapour (in terms of mole fraction) will beA. 0.635 methanol, 0.365 ethanolB. 0.365 methanol, 0.635 ethanolC. 0.574 methanol, 0.326 ehtanolD. 0.173 methanol, 0.827 ethanol |
| Answer» Correct Answer - B | |
| 102. |
The atmospheric pressure is sum of the _______.A. Pressure of the biomoleculesB. Vapour pressure of atmospheric constituentsC. Vapour pressure of chemicals and vapour pressure of volatilesD. Pressure created on to atmospheric molecules |
| Answer» Correct Answer - B | |
| 103. |
The lowering in vapour pressure when a solute is addedA. Will depend upon the nature of the soluteB. Will be the same irrespective of the solvent takenC. Will be different when different solvent are takenD. Will depend on the form ideal solution |
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Answer» Correct Answer - A::C::D Explanation : (a) `Delta P=oP_(A)^(@)x_(B)` (a) Hence, `Delta F` depends upon i which is dependent on nature of solute (Hence, choice (a) is correct). (c ) `Delta F` depends upon `P_(A)^(@)`. If `P_(A)^(@)` changes then `Delta F` changes. Hence when different solvents are taken `Delta F` value is different. Choice (c ) is also correct. (d) Is also correct. `Delta F` changes with mole fraction of solute. `Delta F` also changes with temperature. (b) Is wrong because `Delta F` depends upon solvent. |
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| 104. |
A mixture of two completely miscible non-ideal liquids which distill as such without change in its composition at a constant temperature as though it were a pure liquid. This mixture is known asA. Binary liquid mixtureB. Azeotropic mixtureC. Eutectic mixtureD. Ideal mixture |
| Answer» Correct Answer - B | |
| 105. |
If liquids A and B from an ideal solutionA. The enthalpy of mixing is zeroB. The entropy of mixing is zeroC. The free energy of mixing is zeroD. The free energy as well as the entropy of mixing are each zero |
| Answer» Correct Answer - A | |
| 106. |
Which of the following will form non-ideal solution?A. Benzene + n-heptaneB. n-hexane + n-heptaneC. Ethyl bromide + ethyl iodideD. `C Cl_(4)+CHCl_(3)` |
| Answer» Correct Answer - D | |
| 107. |
Addition of a non-volatile solute in a volatile ideal solventA. Increases the vapoure pressure of the solvent Decreases the vapour pressure of the solventB. Decreases the vapoure pressure of the solventC. Decreases the boililng point of the solventD. Increases the freezing point of the solvent |
| Answer» Correct Answer - B | |
| 108. |
Which of the following modes of expressing concentration is independent of temperatureA. MolarityB. MolalityC. FormalityD. Normality |
| Answer» Correct Answer - B | |
| 109. |
For a non-ideal solution with a-ve deviationA. `Delta H_(mix)=-ve`B. `Delta V_(mix)=-ve`C. `Delta S_(mix)=-ve`D. `Delta G_(mix)=-ve` |
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Answer» Correct Answer - A::B::D Explanation : (a) For -ve deviation `Delta H_(mix)` = -ve due to greater forces of attraction after mixing. Statement (a) is correct. `Delta V_(mix)` = -ve because volume of resulting solution decreases as particles come close together. Statement (b) is correct `Delta G_(mix)=-ve` because `Delta G=Delta H-T Delta S=(-)Delta H-T(+Delta S)=-ve` Hence, choice (d) is correct. `Delta S_(mix) =+ve`, Entropy always increases on mixing. Statement (c )is incorrect. |
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| 110. |
In mixture A and B ,components show -ve deviations as:A. `Delta V_(mix)gt 0`B. `Delta H_(mix)lt 0`C. A-B interaction is weaker than A-A and B-B interactionD. A-B interaction is strong than A-A and B-B interaction |
| Answer» Correct Answer - B::D | |
| 111. |
Osmotic pressure of a solution is `0.0821 atm` at a temperature of 300 K. The concentration in mole/litre will beA. 0.033B. 0.066C. `0.33 xx 10^(-2)`D. 3 |
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Answer» Correct Answer - C `C=pi/(RT) = 0.0821/(0.0821 xx 300) = 1/300 =0.33 xx 10^(-2) M ` |
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| 112. |
The density of `3 M` sodium of thiosulphate solution `(Na_(2)S_(2)O_(3))` is `1.25 g mL^(-1)`. Calculate a. The precentage by weight of sodium thiosulphate. b. The mole fraction of sodium thiosulphate. c. The molalities of `Na^(o+)` and `S_(2)O_(3)^(2-)` ions.A. `12.64%`B. `37.92%`C. `0.87%`D. `63.21%` |
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Answer» Correct Answer - B Mass of 1000 mL of `Na_(2)S_(2)O_(3)` solution `=1.25xx1000=1250 g` Molarity `=("mass")/("molecular mass " xx " volume (mL)")xx100` `rArr 3=("mass " xx 1000)/(158xx1000)` `therefore` Mass of `Na_(2)S_(2)O_(3)` is 1000 mL of 3M solution `=3xx158=474 g` Percentage by weight of `Na_(2)S_(2)O_(3)` in solution `=(474)/(1250)xx100 =37.92 %` |
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| 113. |
NaClO solution reacts with `H_(2)SO_(3)` as, `NaClO+H_(2)SO_(3)rarr NaCl + H_(2)SO_(4)`. A solution of NaClO used in the above reaction contained 15g of NaClO per litre. The normality of the solution would beA. `0.8`B. `0.6`C. `0.2`D. `0.4` |
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Answer» Correct Answer - D `N=(w xx 1000)/(eq.wt. xx "volume in ml.") =0.4 N`. |
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| 114. |
The weight of pure `NaOH` required to prepare `250 cm^(3)` of `0.1N` solution isA. 4 gB. 1 gC. 2 gD. 10 g |
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Answer» Correct Answer - B `N=(w)/(E xx V(l))rArr 0.1=(w)/(40xx0.25)rArr w=1 gm` |
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| 115. |
Hydrolic acid solution A and B have concentration of 0.5 N and 0.1N respectively. The volume of solutions A and B required to make 2 litres of 0.2 N hydrochloric areA. 0.5 L of A + 1.5 L of BB. 1.5 L of A + 0.5 L of BC. 1.0 L of A + 1.0 L of BD. 0.75 L of A + 1.25 L of B |
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Answer» Correct Answer - A Let volume of A = xL Volume of B = (2-x)L `because M_(1)V_(1)+M_(2)V_(2)=M(V_(1)+V_(2))` `rArr 0.5x +0.1(2-x)=0.2(x+2-x)` (`because` for HCl, 0.1 M HCl = 0.1 N HCl) `rArr 0.5 x + 0.2 =0.4 rArr 0.4 x = 0.2` `therefore x =(0.2)/(0.4)=0.5 L` Since x = 0.5 L (vol. of A), then 2 - x = 2 - 0.5 = 1.5 L (Vol. of B). |
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| 116. |
Liquid A and B from ideal solution over the entire range of composition. At temperature T, equimolar binary solution of liquids A and B vapour pressure 45 Torr. At the same temperature, a new solution of A and B having mole fractions `x_(A)` and `x_(B)`, respectively, has vapour pressure of 22.5 Torr. The value of `X_(A)//x_(B)` in the new solution is _________ (Given that the vapour pressure of pure liquid A is 20 Torr at temperature T) |
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Answer» Correct Answer - A `45=P_(A)^(@)xx(1)/(2)+P_(B)^(@)xx(1)/(2)` `P_(A)^(@)+P_(B)^(@)=90` …..(1) Given `P_(A)^(@)=200` torr `P_(B)^(@)=70` torr `rArr 22.5` torr `=20 x_(A)+70(1-x_(A))` `=70-50x_(A)` `x_(A)=((70-22.5)/(50))=0.95` `x_(B)=0.05` So `(x_(A))/(x_(B))=(0.95)/(0.05)=19` . |
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| 117. |
The mole fraction of a solute in its solution in acetic acid is 0.2. The mass of solute (molar mass = 40) in 120g. Of acetic acid would beA. 2 gB. 8 gC. 10 gD. 20 g |
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Answer» Correct Answer - D Mole fraction of acetic acid = 1 - 0.2 = 0.8 :. Solution contains `0.8 xx 60 = 48 `g. of acetic acid and `0.2 xx 40 = 8 ` g of solute :. 120 g acetic acid will contain` (8 xx 120)/48=20 g.` of solute. |
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| 118. |
120g of urea is present in 5L of solution, the active mass of urea isA. 0.2B. 0.06C. 0.4D. 0.88 |
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Answer» Correct Answer - C Active mass = concentration = M ` :. M=W_(2)/(M_(2) xx V (L))=120/(60 xx 5)=12/30=0.4` |
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| 119. |
The mass of oxalic acid crystals `(H_(2)C_(2)O_(4).2H_(2)O)` required to prepare 50 mL of a 0.2 N solution is:A. 126 gB. 12.6 gC. 63 gD. 6.3 g |
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Answer» Correct Answer - D 1000 ml of 1 N oxalic solution = 63 g 500 ml of 0.2 N oxalic acid solution `=(63)/(1000)xx500xx0.2=6.3 g`. |
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| 120. |
The mole fraction of oxalic acid (Molar mass 63) required to prepare 0.10 m solution in water isA. 1B. 0.0018C. 6.3D. 0.0992 |
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Answer» Correct Answer - B `X_(2)= (mM_(1))/(1000+mM_(1))` `=(0.1 xx 18)/(1000 + 0.1 xx 18)` =0.0018 |
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| 121. |
What weight of hydrated oxalic acid should be added for complete neutralisation of 100 ml of 0.2N - NaOH solutionA. 0.45 gB. 0.90 gC. 1.08 gD. 1.26 g |
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Answer» Correct Answer - D For complete neutralization equivalent of axalic acid = equivalent of NaOH = `(w)/(eq.wt)=(NV)/(1000) therefore (w)/(63)=(0.2xx100)/(1000)rArr w=1.26 gm`. |
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| 122. |
The number of molecules in 16 g of methane isA. `3.0xx10^(23)`B. `6.02xx10^(23)`C. `(16)/(6.02)xx10^(23)`D. `(16)/(3.0)xx10^(23)` |
| Answer» Correct Answer - B | |
| 123. |
With 63 gm of oxalic acid how many litres of `(N)/(10)` solution can be preparedA. 100 litreB. 10 litreC. 1 litreD. 1000 litre |
| Answer» Correct Answer - B | |
| 124. |
How many grams of glucose be dissolved to make one litre solution of `10%1` glucose:A. 100 gmB. 180 gmC. 18 gmD. 1.8 gm |
| Answer» Correct Answer - A | |
| 125. |
The molarity of a solution containing `5.0g` of NaOH in 250 mL solution is :A. 0.1 MB. 1 MC. 0.01 MD. 0.001 M |
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Answer» Correct Answer - A `M=(w xx1000)/(m xx "Volume in ml.")=(1xx1000)/(40xx250)=0.1 M`. |
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| 126. |
Pick out the wrong statement(s) 1. Vapour pressure of a liquid is the measure of the strength of intermolecular attractive forces 2. Surface tension of a liquid acts perpendicular to the surface of the liquid 3. Vapour pressure of all liquids is same at their freezing points 4. Liquids with stronger intermolecular attractive forces are more visscous than those with weaker intermolecular forcesA. 2,3 and 4B. 2 and 3C. 1, 2 and 3D. 3 only |
| Answer» Correct Answer - D | |
| 127. |
Define molal elevation constant or ebullioscopic constant.A. `.^(@)C//m`B. K/mC. K kg `mol^(-1)`D. K mol `kg^(-1)` |
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Answer» Correct Answer - D (d) is not the correct unit of `K_(b)` |
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| 128. |
On mixing 10 mL of acetone with 50 mL of chloroform, the total volume of the solution isA. `lt 60 mL`B. `gt 60 mL`C. `= 60 mL`D. unpredictable |
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Answer» Correct Answer - a Chloroform and acetone show strong hydrogen bonding. Hence, for such solution `(DeltaV_(mix) = -ve)`, volume decreases on mixing (Non-ideal solution with negative deviation). |
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| 129. |
On mixing 25 mL of `C CL_(4)` with 25 mL of toluene, the total volume of the solution isA. `= 50 mL`B. `gt 50 mL`C. `lt 50 mL`D. unpredictable |
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Answer» Correct Answer - b It is a non-ideal solution, hence, `DeltaV_(mix) = +ve` |
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| 130. |
`MX_(2)` dissociates into `M^(2+)` and `X^(-)` ions in an aqueous solution, with a degree of dissociation `(alpha)` of 0.5. The ratio of solution to the value of depression of freezing point in the absence of ionic dissociation is |
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Answer» Correct Answer - 2 `MX_(2)hArr M^(2+)+2X^(-)` `m_(0)(1-alpha) " " m_(0)alpha " " 2m_(0)alpha , m=m_(0)(1+2alpha)` `therefore m=m_(0)(1+2xx0.5)=2m_(0)` (as given) `((-Delta T_(f))_("observed"))/((-Delta T_(f))_("undissociated"))=i=(m)/(m_(0))=2` |
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| 131. |
In a 0.2 molal aqueous solution of a week acid HX, the degree of ionisation is 0.3. Taking `K_(F)` for water as 1.85, the freezing point of the solution will be nearest toA. `-0.360^(@)C`B. `-0.260^(@)C`C. `+0.480^(@)C`D. `-0.481^(@)C` |
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Answer» Correct Answer - d `{:(,HX, hArr,H^(+),+,X^(-)),("Initially",1 mol,,0,,0),("After dissociation",1-0.3,,0.3,,0.3):}` Total moles `= 1 - 0.3 + 0.3 + 0.3 = 1.3` `I = (1.3)/(1) = 1.3` `DeltaT_(f) = iK_(f)M = 1.3xx1.85xx0.2 = 0.481^(@)C` |
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| 132. |
8 g of HBr is added in 100 g of `H_(2)`The freezing point will be `(K_(f) = 1.86, H=1, br = 80 )`A. `-0.75^(@)C`B. `0^(@) `C. `-3.67 ^(@) C`D. `-7.6 ^(@) C ` |
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Answer» Correct Answer - C `Delta T_(f) = iK_(f) xx (W_(2) xx 1000)/(M_(2) xx W_(1)) = 2 xx 1.86 xx (8 xx 100)/(81 xx 100)=3.67^(@)` `T_(f) = T^(@) - Delta T_(f) = -3.67^(@) C` |
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| 133. |
Pure benzene freezes at `5.45^(@)C` at a certain place but a 0.374m solution of tetrachloroethane in benzene freezes at `3.55^(@)C`. The `K_(f)` for benzene isA. 5.08 K kg `"mol"^(-1)`B. 508 K kg `"mol"^(-1)`C. 0.508 K kg `"mol"^(-1)`D. `50.8^(@) C `kg `"mol"^(-1)` |
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Answer» Correct Answer - A `K_(f) = Delta T_(f) // m = (5.45 - 3.55) // 0.374` `= 5.08 K kg "mol"^(-1)` |
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| 134. |
The freezing point of 0.1 M solution of glucose is `-1.86^(@)C`. If an equal volume of 0.3 M glucose solution is added, the freezing point of the mixture will beA. `-7.44^(@)C`B. `-5.58^(@)C`C. `-3.72^(@)C`D. `-2.79^(@)C` |
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Answer» Correct Answer - C Freeing point of mixture is `M_(1)V_(1) + M_(2)V_(2) = M_(3)V_(3)` `K_(f) = Delta T_(f) // m = 1.86 //0.1 = 18.6` `0.1 xx V + 0.3 xx V = M_(3) xx 2V` or `M_(3) = 0.2, Delta T_(f) = 18.6 xx 0.2 = 3.72^(@)` `T_(f) = 3.72^(@) C` |
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| 135. |
The freezing point of 0.1 M solution of glucose is `-1.86^(@)C`. If an equal volume of 0.3 M glucose solution is added, the freezing point of the mixture will beA. `-7.44^(@)C`B. `-5.58^(@)C`C. `-3.72^(@)C`D. `-2.79^(@)` |
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Answer» Correct Answer - c `K_(f) = DeltaT_(f)//M = 1.86//0.1 = 18.6` `0.1xxV + 0.3 V = M_(3)xx2V` or `M_(3) = 0.2` `DeltaT_(f) = 18.6xx0.2 = 3.72^(@)C` `T_(f) = 18.6xx0.2 = 3.72^(@)C` `T_(f) = -3.72^(@)C` |
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| 136. |
40% by weight solution will contain how much mass of the solute in 1 solution, density of the solution is 1.2 g/mLA. 480 gB. 48 gC. 38 gD. 380 g |
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Answer» Correct Answer - A Molarity `= ("% by weight of solute" xx "density of solution" xx 10"(in litre)")/(M)` when M = mol. Weight of the solute Molarity `=(40xx1.2xx10)/(Mxx1000)` .......(i) Molarity `=("weight of the solute / M")/("volume of solution (in litre)")`. .........(ii) From Eqs. (i) and (ii) `("Weight of solute")/(M xx 1000)=(40xx1.2xx10)/(M xx 1000)` Weight of solute = 480 g. |
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| 137. |
A `x` molal solution of a compound in benzene has mole fraction of solute equals to 0.2. The value of `x` isA. 14B. `3.2`C. 4D. 2 |
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Answer» Correct Answer - B `(X)/(X + (1000)/(78))=0.2` |
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| 138. |
An X molal solution of a compound in benzene has mol e fraction of solute ` =0.2. ` The value of X isA. 14B. 3.2C. 1.4D. 2 |
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Answer» Correct Answer - b Molality and mole fractions of solute `(x_(A))` are related as `x_(A) = (m.M_(B))/(1000+m.M_(B)) rArr 0.2 = (mxx78)/(1000+mxx78)` `200 + 15.6 m = 78 m` or 62.4 m = 200 m = 3.2 |
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| 139. |
Out of molarity (M), molality (m), normality (N) and mole fraction (x), those independent of temperature areA. M and mB. N and xC. m and xD. M and x |
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Answer» Correct Answer - c Molality (m) and mole fraction (x) involves only mass and do not depend upon volume. Hence, these are independent of temperature. |
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| 140. |
Define the term solution. How many types of solutions are formed ? Write briefly about each type with an example. |
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Answer» A solution is homogeneous mixture of two or more chemically non-reacting substnaces. Types of solutions. There are nine types of solution types of solution examples. Gaseous solution (a) gas in gas air, mixture of `O_(2)` and `N_(2)`. Etc. (b). Liquid in gas water vapour. (c) solid in gas comphor vapours in `N_(2)` gas smoke etc. liquid solution (a). Gas in liquid `CO_(2)` dissolved in water (aerated water). and `O_(2)` dissolved in water etc. (b). Liquid in liquid ethanol dissolved in water etc. (c). Solid in liquid sugar dissolved in water, etc Solid solutions (a). Gas in solid solution of hydrogen in palladium (b). liquid in solid amalgams e.g., Na-Hg (c). Solid in solid gold ornamens (Cu/Ag with Au). |
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| 141. |
Each question contains STATEMENT-I(Assertion) and STATEMENT-2(Reason).the statement carefully and mark the correct answer accoring to the instrution given below: STATEMENT - 1 : The molecular mass of acetic acid determined by depression in freezing point method in benzene and water was found to be differrent. STATEMENT - 2 : Water is polar and benzene is non-polar.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false. |
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Answer» Correct Answer - A Depression in freezing point is a colligative property which depends upon the number of particles are different in case of benzene and water that is why molecular weight of acetic acid determined by depression in freezing point method is also different. |
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| 142. |
Desalination of sea water can be done byA. OsmosisB. Reverse osmosisC. FiltrationD. Diffusion |
| Answer» Correct Answer - BB. Reverse Osmosis | |
| 143. |
An aqueous solution of methanol in water has vapour pressureA. Equal to that of waterB. Equal to that of methanolC. More than that of waterD. Less than that of water |
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Answer» Correct Answer - C Methanol has low boiling point than `H_(2)O` Lower is boiling point of solvent more is vapour pressure. |
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| 144. |
Statement 1 : Lowering of vapour pressure is directly proportional to the number of species present in the solution. Statement 2 : The vapour pressure of 0.1M sugar solution is more than 0.1 M KCl solution.A. Statement 1 is true, statement 2 is true, statement 2 is a correct explanation for statement 2B. Statement 1 is true, statement 2 is true , statement 2 is not a correct explanation for statement 2C. Statement 1 is true, statement 2 is falseD. Statement 1 is false, statement 2 is true |
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Answer» Correct Answer - A Lowering of vapour pressure is a colligative property. |
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| 145. |
Vapour pressure of pure A = 100 torr, moles = 2. Vapour pressure of pure B = 80 torr, moles = 3. Total vapour pressure of mixture isA. 440 torrB. 460 torrC. 180 torrD. 88 torr |
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Answer» Correct Answer - D `P_("total")=p_(A)^(@)x_(A)+p_(B)^(@)x_(B)` `P_("total")=P_(A)+P_(B)=100xx(2)/(5)+80xx(3)/(5)=88` torr. |
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| 146. |
The vapour pressure of water at `20^(@)C` is 17.54 mm. When 20 g of a non-ionic substance is dissolved in 100g of water, the vapour pressure is lowered by 0.30 mm . What is the molecular weight of the substancesA. `210.2`B. `206.88`C. `215.2`D. `200.8` |
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Answer» Correct Answer - B `(P_(o)-P_(s))/(P_(s))=(w)/(m)xx(M)/(W)` `P_(o)=17.54 mm` `P_(s)=17.24 mm` M = 18 W = 100 w = 20 m = ? `(17.54-17.24)/(17.24)=(20)/(m)xx(18)/(100)` `(30)/(17.24)=(360)/(100m)` `m=(360xx17.24)/(100xx30)rArr206.88` |
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| 147. |
Statement 1 : Entropy of solution is less than entropy of pure solvent. Statement 2 : The freezing point of water is depressed by the addition of glucose.A. Statement 1 is true, statement 2 is true, statement 2 is a correct explanation for statement 1B. Statement 1 is true, statement 2 is true , statement 2 is not a correct explanation for statement 1C. Statement 1 is true, statement 2 is falseD. Statement 1 is false, statement 2 is true |
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Answer» Correct Answer - D `Delta T_(f)=mK_(f)i` Entropy of solution is greater than entropy of solvent due to presences of greatere number of particles in solution. |
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| 148. |
Vapour pressure increases with increase inA. Concentration of solution containing non-volatile soluteB. Temperature upto boiling pointC. Temperature upto triple poitnD. Altitude of the concerned place of boiling |
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Answer» Correct Answer - B Vapour pressure becomes identical as the atmospheric pressure at boiling point. If the liquid is heated beyond that only evaporation continues, vapour pressure does not rise further. |
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| 149. |
Which of the following is not correct for `D_(2)O`A. Boiling point is higher than `H_(2)O`B. `D_(2)O` reacts slowly than `H_(2)O`C. Viscosity is higher than `H_(2)O` at `25^(@)`D. Solubility of NaCl in it is more than `H_(2)O` |
| Answer» Correct Answer - D | |
| 150. |
Dust is an example of-A. solid solutionB. liquid solutionC. gas solutionD. none |
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Answer» Correct Answer - C Dust is the particles in air |
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