MCQ Questions for Class 12 Engineering Chemistry Solutions Quiz 1

When

$200\ g$ of an oleum sample labelled as

$109\%$ is mixed with

$300\ g$ of another oleum sample labelled as

$118\%$ the new labelling of resulting oleum sample becomes

0%
$114.4\%$
0%
$112.6\%$
0%
$113.5\%$
0%
$127\%$

Explanation

$200\ gm$ of

$109\%$ oleum sample

$300\ gm$ of

$118\%$ oleum sample

$SO_3+H_2O\to H_2SO_4$

$80\ gm\ 18\ gm\ 28\ g$
In

$109\%$ labelled sample

$9\ gm\ H_2O$ react with

$\dfrac {80}{18}\times 9=40\ gm\ SO_3$

$\therefore \ 40\%\ SO_3+60\% \ H_2SO_4$

$\therefore \$ In

$200\ gm$ sample

$80\ gm\ SO_3$

$12\ gm\ H_2SO_4$
Now
In

$118\%$ sample

$18\ gm\ H_2O$ react with

$80\ gm\ SO_3$

$\therefore \ 80\% \ SO_3\ ;\ 20\% \ H_2SO_4$

$\therefore \$ In

$300\ gm$ sample :-

$240\ gm\ SO_3$

$60\ gm\ H_2SO_4$
When both sample mined:-

$SO_3 :- 80+240=320\ gm\ H_2SO_4 :- 120+60=180\ gm$

$\therefore \ \%$ of

$SO_3=\dfrac {320}{320+180}\times 100=64\%$

$64\ gm\ SO_3$ combine with :-

$\dfrac {18}{80}\times 64=14.4\ gm\ H_2O$

$\therefore \$ New label of resulting sample

$=100+14.4$

$=114.4\%$

A sample of oleum is labelled as

$112\%$ . In

$200\ g$ of this sample,

$18\ g$ water is added, The resulting solution will contain

0%
$218\ g$ pure $H_2SO_4$
0%
$218\ g\ H_2SO_4$ and $6\ g$ free $SO_3$
0%
$212\ g\ H_2SO_4$ and $6\ g$ free $SO_3$
0%
$191.33\ g\ H_2SO_4$ and $26.67\ g$ free $SO_3$

Explanation

$200\ g$ sample of oleum labelled

$112\%$

$SO_3+H_2O\to H_2SO_4$

$80\ gm\ 18\ gm\ 98\ gm$

$112\%$ sample

$12\ gm\ H_2O$ react with

$\dfrac {80}{18}\times 12$

$=53.33\ gm\ SO_3$

$\therefore \ \% SO_3=53.33\%$ (free

$SO_3$ )

$\%\ H_2SO_4=46.67\%$
In

$200\ mg$ sample

$SO_3=106.66\ gm$

$H_2SO_4=93.34\ gm$
Now

$18\ g\ H_2O$ is added

$\therefore \ SO_3$ converted to

$H_2SO_4=80\ gm$
and from

$98\ gm\ H_2SO_4$
Now

$SO_3=106.66-80=26.66\ gm\ SO_3$

$H_2SO_4=93.34+98=191.34\ gm\ H_2SO_4$

Vapour pressure of liquid

0%
increases with increase in temperature
0%
decrease with increase in temperatuyre
0%
is independent of temperature
0%
either increases or decreases with the increase in temperature, depending on the nature of liquid

Which of the following behaviour is true about the ideal binary liquid solution of liquids A and B if

$P^0_A < P^0_B$ ?

0%
plot of $P_{total}| vs X_A$ in non linear
0%
plot of $P_{total}| vs X_B$ is linear with +ve slope
0%
plot of $P_{total}| vs X_B$ is linear with slope = 0
0%
plot of $P_{total}| vs X_B$ is linear with -ve slope

Benzene and toulene from an ideal solution, the vapour pressures of the benzene and toluene are 75 mm and 25 mm respectively at

$20^0 C$ if the mole fractions of benzene and toluene in vapour are 0.75 and 0.25 respectively the vapour pressure of the ideal solution is

0%
62.5 mm
0%
50 mm
0%
30 mm
0%
40 m

The vapour pressure of a soluiton of two liquids A (

$P^0 = 80 mm , X = 0.4 )$ and

$B ( P^0 = 120mm ,X=0.6 )$ is found to be 100 mm, it shows that the solution exhibits

0%
negative deviation from ideal behaviour
0%
positive deviation from ideal behaviour
0%
ideal behaviour
0%
positive deviation at lower concentration

Heptane and octane from ideal solution at 373 K the vapour pressure of the pure liquids are 106 kPa and 46 kpa respectively . what will be the vapour pressure, in bar of mixture of 30.0 g of heptane and 34.2 g of octane ?

0%
76 bar
0%
152 bar
0%
1.52 bar
0%
0.76 bar

For an ideal solution of A and B ,

$Y_A$ is the mole fraction of A in the vapour phase at equilibrium . which of the following plot should be linear ?

0%
$P_{total} vs Y_A$
0%
$P_{total} vs Y_B$
0%
$\frac {1}{P_{total } } vs Y_A$
0%
$\frac {1}{P_{total} } vs \frac {1}{Y_A }$

Under the condition of a similar temperature, which of the following solution will have, minimum vapour pressure?

0%
0.1 M sugar
0%
0.1 M $NaCl$
0%
0.1 m $BaCl_2$
0%
0.1 M $Al_2(SO_4)_3$

A mixture contains 1 mole of volatile liquid

$A (P^0_A = 100 mm Hg )$ and 3 moles of volatile liquid

$B (P^0_B = 80 mm Hg )$ if the solution behaves ideally, the total vapour pressure of the distillate is

0%
85 mm Hg
0%
85.88 mm Hg
0%
90 mm Hg
0%
92 mm Hg

Explanation

Mole fraction of A is

$X_A=\frac{1}{1+3}=0.25$

Mole fraction of B is

$X_B=1-0.25=0.75$

The expression for the total pressure of the solution is

$P=P_A^0X_A+P_B^0X_B$

Substitute values in the above expression

$P=100\times 0.25+80\times 0.75=85$ mm of Hg

Hence the correct answer is [A]

An aqueous solution of sucrose is 0.5 molal. What is the vapour pressure of water above this solution ? The vapour pressure of pure water is 25.0 mm Hg at this temperature.

0%
24.8 mm Hg
0%
0.45 mm Hg
0%
2.22 mm Hg
0%
20.3 mm Hg

Explanation

Given that

$m = 0.5\;\;molal$

V.P. (pure water) 25.0mm Hg

According to the Raoult's law for dilute solution

$\dfrac{P_0 - P_s}{P_0} = \dfrac{w}{m+W}\times 1000 = X_{sucrose}$

here the 0.5 molal of solution means 0.5 mole in 1kg of water

Henc the mole of solvent =

$\dfrac{1000}{18}=55.5$ [ the molecular mass of water molecules is 18]

then

$\dfrac{25-P_s}{25} = \dfrac{0.5}{55.57+0.5}$

$\Rightarrow\;1401.75 - 56.07P_s = 12.5$

$\Rightarrow \; 56.07P_s = 1407.75 - 12.5$

$\Rightarrow \; P_S = \dfrac{1389.25}{\;\;56.07}$

$24.8\;mm \;Hg$

A liquid mixture of A and B (assume ideal solution ) is placed in a cylinder and piston arrangement . The piston is slowly pulled out isothermally so that the volume of liquid decreases and that of vapour increases. at the instant when the quantity of the liquid still remaining is negligibly small, the mole fraction of 'A' in the vapour is 0.4 .If

$P^0_A = 0.4 atm , P^0_B = 1.2 atm$ at this temperature, the total pressure at which the liquid has almost evaporated is ?

0%
0.667 atm
0%
1.5 atm
0%
0.8 atm
0%
0.545 atm

Liquids A and B from an ideal solution. A certain solution of A and B contains 25 mole percent of A whereas the vapours in the equilibrium with the solution at 298 K contains 50 mole per cent of A. The ratio of vapour pressures of pure A to that of pure B at 298 K , is

0%
1 :1
0%
3 : 1
0%
1 :3
0%
2 :1

The ratio between lowering of vapour pressure of solution and mole fraction of solute is equal to

0%
relative lowering of vapour pressure
0%
vapour pressure of pure solvent
0%
vapour pressure of solution
0%
molar mass of solvent

For each of the following dilute solutions van't Hoffs factor is equal of 3 except

0%
$Na_2SO_4$
0%
$CaF_2$
0%
$K_3PO_4$
0%
$(NH_4)_2CO_3$

Liquids A and B from an ideal solution. the plot of

$\dfrac {1}{X_A} (y-axis) \,vs \,\dfrac {1}{Y_A }(x\, axis )$ is linear whose slope and intercept , respectively, are given as

0%
$\dfrac {P^0_A }{ P^0_B} , \dfrac { P^0_A - P^0_B}{P^0_B}$
0%
$\dfrac {P^0_A }{ P^0_B} , \dfrac { P^0_B - P^0_A}{P^0_B}$
0%
$\dfrac {P^0_B }{ P^0_A} , \dfrac { P^0_A - P^0_B}{P^0_B}$
0%
$\dfrac {P^0_B }{ P^0_A} , \dfrac { P^0_A - P^0_B}{P^0_A}$

Explanation

The plots

$\frac{1}{X_A}$ vs

$\frac{1}{Y_A}$ where

$X_A$ and

$Y_A$ are the mole fraction of liquid A in liquid and vapour phase respectively) is linear with slope and intercepts equal to

$P_{A}^{0}\setminus P_{B}^{0}$ and

$\frac{P_B^0-P_A^0}{P_B^0}$ respectively

$P_{A}^{'}=P_{A}^{0}\times X_{A}$ and

$P_{B}^{'}=P_{B}^{0}\times X_{B}$

$P_{A}^{'}=P_{M}\times Y_{A}$ and

$P_{B}^{'}=P_{M}\times Y_{B}$

$\therefore \frac{P_A}{Y_A}=\frac{P_B}{Y_B}$

$\frac{P_A^0\times X_A}{Y_A}=\frac{P_B^0\times X_B}{Y_B}=\frac{P_B^0 \times (1-X_A)}{1-Y_A}$

$\frac{P_B^0}{X_A}=\frac{P_A^0}{Y_A}+(P_B^0-P_A^0)$

$\frac{1}{X_A}=\frac{1}{Y_A}\times \frac{P_A^0}{P_B^0}+\frac{P_B^0-P_A^0}{P_B^0}$

$y=mx+C$

$\therefore Slope=m=\frac{P_A^0}{P_B^0}$ and Intercept

$C=\frac{P_B^0-P_A^0}{P_B^0}$

Hence the correct answer is [B}.

The limiting value of van't Hoff's factor for

$Na_2SO_4$ is

0%
2
0%
3
0%
4
0%
5

At what pressure half of the total moles of liquid solution will vaporize ?

0%
567.14 mm Hg
0%
63.25 mm Hg
0%
70 mm Hg
0%
66.67 mm Hg

If the pressure over the mixture at 300 K is reduced at what pressure does the first vapour form?

0%
40 mm Hg
0%
70 mm Hg
0%
100 Mm Hg
0%
199 mmHg

A liquid solution is formed by mixing 10 moles of anline and 20 moles of phenol at a temperature where the vapour phenol are 90 and 897 mm Hg, respectively . The possible vapour pressure of solution at that temperature is ?

0%
82 mm Hg
0%
88 mm Hg
0%
90 mm Hg
0%
94 mm Hg

If the pressure is reduced further, at what presure does the trace of liquid disppear?

0%
57.14 mm Hg
0%
40 Mm Hg
0%
100 mm Hg
0%
66.67 mm Hg

The vapour pressure of pure liquids A, B and C are 75, 22 and 10 torr , respectively. Which of the following is /are possible values of vapour pressure of binary or ternary solutions having equimolar amounts of these liquids? Assume ideal behaviour for all possible solutions

0%
53.5 torr
0%
35.67 torr
0%
48.5 torr
0%
16 torr

Explanation

$P_{AB}= \cfrac{75+22}{2}=48.5$ torr

$P_{BC}= \cfrac{22+10}{2}=16$ torr

$P_{AC}= \cfrac{75+10}{2}=42.5$ torr

$P_{ABC}= \cfrac{75+22+10}{2}=35.67$ torr

Option A, B and C are correct.

The vapour pressure of the solution this composition is:

0%
$\sqrt { P^0_A .P^0_B }$
0%
$( P^0_A - P^0_B )$
0%
$( P^0_A + P^0_B )$
0%
$0.5 ( P^0_A + P^0_B )$

Which of the following is the only incorrect informarion regarding the composition of the system at 0.51 bar pressure ?

0%
$X_A = 0.45$
0%
$Y_A = \frac {6}{17}$
0%
$n_{A(liquid) } = \frac {12}{11}$
0%
$n_{A(vapour) } = \frac {12}{11}$

If the vapour are compressed slowly and isothermally , at what pressure , the first drop of liquid will appear ?

0%
0.4 bar
0%
0. 5 bar
0%
0.52 bar
0%
0.6 bar

In a binary electrolyte (AB type ) the observed lowering of vapour pressure as compared to the theorotically calculated one for non electrolyte can never be

0%
more than double
0%
exactly equal to double
0%
less than double
0%
more than one

Explanation

i = 1 +

$\alpha (n-1)$

As AB is a binary electrolyte hence the value of n = 2

$\therefore$ i = 1 +

$\alpha$

For observed lowering of vapour pressure

$\alpha$ can not be equal to 1 it should be always less than 1.

$\dfrac{P^0-P}{P^0} = ix_B$

$\therefore$ The observed lowering of vapour pressure as compared to the theoretically calculated one for binary electrolyte can never be more than double and exactly equal to double as

$1 < i < 2$ .

One having high vapour pressure at temperature below its melting point is?

0%
Benzoic acid
0%
Salicylic acid
0%
Citric acid
0%
All of these

Explanation

Benzoic acid appears as a white crystalline solid. Vapour pressures of benzoic acid were measured in the temperature range

$316\,K$ to

$391 \,K$ which is below its melting point

$395.1\, K$ . Hence correct answer is option A.

The vapour pressure of pure liquid A at

$80^0 C$ is

0%
807.4 mm
0%
511.1 mm
0%
755.6 mm
0%
533.3 mm

Two liquids A and B form an ideal solution. the solution has a vapor pressure of 700 torrs at

$80^0 C$ . it is distilled till 2/3rd of the solution is collected as condensate. the composition of the condensate

$x'_A = 0.75$ . and that of residue is

$x''_A = 0.30$ if the vapor pressure of the residue at

$80^0 C$ is 600 torr. which of the following is/are true?

0%
The composition of the original liquid was $x_A = 0.6$
0%
The composition of the original liquid was $x_A = 0.4$
0%
$P^0_A = \frac {2500}{3} Torr$
0%
$P^0_B = 500 Torr$

Explanation

$x_AP^0_A + x_B P^0_B = 700 ......(i)$

$X_A "P^0_A + X^"_B P^0_B = 0.30 P^0_A + 0.70 P^0_B = 600 .....(ii)$
If moles of A ^b initially are C & Y then

$X = 0.75 \times \frac {2}{3} (x +y) + 0.30 \times \frac {1}{3}( x +y)......(iii)$

$x_A = \frac {X}{x +y} \quad or \quad x_B = \frac {y}{x +y} ......(iv)$
solving gives

$X_A = 0.6 \quad x_B = 0.4 , P^0_A = \frac { 2500}{3} torr$ &

$P^0_B = 500 torr$

The reading of pressure gauge at which only liquid phase exists

0%
499
0%
399
0%
299
0%
none

Explanation

$X_A = \dfrac34=0.75; \quad X_B = 0.25$

$P_{\text{bubble point}} = X_A P^0_A + X_B P^0_B$

$0.75 +400 + 0.25 \times 800 = 500 mm$

Above 500 mm Hg only liquid phase exists.

$y_A = 0.75; \quad \quad y_B = 0.5$

At dew point:

$\dfrac {1}{P_T } = \dfrac {Y_A }{P^0A} + \dfrac {Y_B}{P^0_B } \Rightarrow \dfrac {1}{ P_T} = \dfrac { 0.75}{400} + \dfrac { 0.25}{800} = \dfrac { 1.5 + 0.25}{800}$

$\Rightarrow P_T = \dfrac {800}{1.75} = 457.14\ mm Hg$

Below the dew point, only the vapour phase exists.

On mixing heptane and octane from an ideal solution. At

$373 K$ , the vapour pressure of the two liquid compounds (heptane and octane) are

$105\ kPa$ and

$45\ kPa$ respectively. Vapour pressure of the solution obtained by mixing

$250g$ of heptane and

$35g$ of octane will be (molar mass of heptane

$=100$ and octane

$=114\ g\ mol^{-1}$ )

0%
$72.0\ kPa$
0%
$36.1\ kPa$
0%
$98.4\ kPa$
0%
$114.5\ kPa$

Explanation

Moles of Heptane =

$\mathrm{\cfrac{250}{100} =2.5}$

Moles of Octane =

$\mathrm{\cfrac{35}{114} = 0.3}$

Mole fraction of Heptane =

$\mathrm{\cfrac{2.5}{2.8} = 0.89}$ and that of Octane =

$\mathrm{\cfrac{0.3}{2.8} = 0.11}$

According to Raoult's Law:

$\mathrm{P_T =P^0_A X_A +P^0_BX_B}$

$\mathrm{\implies P_T = (0.89 \times 105 + 0.11 \times 45)}$ kPa

$\mathrm{\implies P_T = 98.4}$ kPa

Which of the following is true when components forming an ideal solution are mixed?

0%
$\Delta H_m = \Delta V_m = 0$
0%
$\Delta H_m > \Delta V_m$
0%
$\Delta H_m < \Delta V_m$
0%
$\Delta H_m = \Delta V_m = 1$

Explanation

For the ideal solution

$\Delta H_{mix}$ and

$\Delta V_{mix} = 0.$

Which pair from the following will not form an ideal solution?

0%
$CCl_4 + SiCl_4$
0%
$H_2O + C_4H_9OH$
0%
$C_2H_5Br + C_2H_5I$
0%
$C_6H_{14} + C_7H_{16}$

Explanation

$H_2O$ and

$C_4H_9OH$ do not form ideal solution because there is hydrogen bonding between

$H_2O$ and

$C_4H_9OH.$

$60 \,gm$ of Urea (Mol. wt 60) was dissolved in

$9.9$ moles of water. If the vapour pressure of pure water is

$P_0$ , the vapour pressure of solution is

0%
$0.10 \,P_0$
0%
$1.10 \,P_0$
0%
$0.90 \,P_0$
0%
$0.99 \,P_0$

Explanation

Given that,

Moles of water.

$N=9.9$ moles

Moles of urea,

$n=\dfrac{w}{M}=\dfrac{60}{60}=1$

Now,

$\dfrac {P^{0} - P_{S}}{P^{0}} = \dfrac {n}{N} \Rightarrow \dfrac {P^{0} - P_{S}}{P^{0}} = \dfrac {1}{9.9} \Rightarrow 9.9 \,P^{0} - 9.9 \,P_s = P^{0}$

$\Rightarrow 8.9 \,P^{0} = 9.9 \,P_s \Rightarrow P_s = \dfrac {8.9}{9.9} P^{0} \approx 0.90 \,P^{0}$

The vapour pressure lowering caused by the addition of

$100 \,g$ of sucrose (molecular mass = 342) to

$1000 \,g$ of water if the vapour pressure of pure water at

$25 ^\circ C$ is

$23.8 \,mm \,Hg$

0%
$1.25 \,mm \,Hg$
0%
$0.125 \,mm \,Hg$
0%
$1.15 \,mm \,Hg$
0%
$00.12 \,mm \,Hg$

Explanation

Given molecular mass of sucrose

$= 342$

Moles of sucrose

$= \dfrac {100}{342} = 0.292 \,mole$

Moles of water

$N = \dfrac {1000}{18} = 55.5 \,moles$ and

Vapour pressure of pure water

$P^{0} = 23.8 \,mm \,Hg$

According to Raoult's law

$\dfrac {\Delta P}{P^{0}} = \dfrac {n}{n + N} \Rightarrow \dfrac {\Delta P}{23.8} = \dfrac {0.292}{0.292 + 55.5}$

$\Delta P = \dfrac {23.8 \times 0.292}{55.792} = 0.125 \,mm \,Hg$

$5 \,cm^3$ of acetone is added to

$100 \,cm^3$ of water, the vapour pressure of water over the solution

0%
It will be equal to the vapour pressure of pure water
0%
It will be less than the vapour pressure of pure water
0%
It will be greater than the vapour pressure of pure water
0%
It will be very large

Which property is shown by an ideal solution

0%
It forms Raoult's law
0%
$\Delta H_{mix} = 0$
0%
$\Delta V_{mix} = 0$
0%
All of these

Explanation

Ideal Solutions generally have characteristics as follows:

$\text{They follow Raoult’s Law.}$
$\text{The enthalpy of mixing of two components should be zero, that is, } Δmix H = 0.$ This signifies that no heat is released or absorbed during mixing of two pure components to form an ideal solution
$\text{The volume of the mixing is equal to zero that is, ΔmixV = 0.}$ This means that total volume of solution is exactly same as the sum of the volume of solute and solution. Adding further, it also signifies that there will be contraction or expansion of the volume while the mixing of two components is taking place.
The solute-solute interaction and solvent-solvent interaction is almost similar to the solute-solvent interaction.

Which has maximum vapour pressure?

0%
$HI$
0%
$HBr$
0%
$HCl$
0%
$HF$

Explanation

The lower is boiling point more is the vapour pressure. Boiling point order is

$HCl < HBr < HI < HF$

An example of near ideal solution is

0%
n-heptane and n-hexane
0%
$CH_3COOH + C_5H_5N$
0%
$CHCl_3 + (C_2H_5)_2O$
0%
$H_2O + HNO_3$

Vapour pressure of a solution is

0%
Directly proportional to the mole fraction of the solvent
0%
Inversely proportional to the mole fraction of the solute
0%
Inversely proportional to the mole fraction of the solvent
0%
Directly proportional to the mole fraction of the solute

Which of the following solution in water possesses the lowest vapour pressure?

0%
$0.1 (M) NaCl$
0%
$0.1 (M) BaCl_2$
0%
$0.1 (M) KCl$
0%
None of these

Which of the following mixture shows positive deviation by ideal behaviour

0%
$CHCl_3 + (CH_3)_2CO$
0%
$C_6H_6 + C_6H_5CH_3$
0%
$H_2O + HCl$
0%
$CCl_4 + CHCl_3$

Explanation

$A. CHCl_3+ (CH_3)_2O$

Presence of strong H-bonding leading negative deviation from Raoult's law.

$(B). C_6H_6+ C_6H_5CH_3\Rightarrow ideal \ solution.$

$(C). H_2O+HCl \Rightarrow$ Water and hydrochloric acid form miscible solutions. They show no deviation

$(D) CCl_4+CHCl_3$ due to presence of Difference in polarity they exhibit +ve deviation from Raoult's law

Which property is not found in ideal solution

0%
$P_A \neq P^0_A \times X_A$
0%
$\Delta H_{mix} \neq 0$
0%
$\Delta V_{mix} \neq 0$
0%
All of these

Explanation

Ideal Solutions generally have characteristics as follows:

$\text{They follow Raoult’s Law.}$ This implies that the partial pressure of components A and B in a solution will be $P_A = P_A^0\times X_A$ .
The enthalpy of mixing of two components should be zero, that is, $Δmix H = 0.$ This signifies that no heat is released or absorbed during mixing of two pure components to form an ideal solution
The volume of the mixing is equal to zero that is, $Δmix V = 0.$ This means that the total volume of solution is exactly the same as the sum of the volume of solute and solution. Adding further, it also signifies that there will be contraction or expansion of the volume while the mixing of two components is taking place.
The solute-solute interaction and solvent-solvent interaction are almost similar to the solute-solvent interaction.

thus, the correct answer is D.

Two

$5$ molal solutions are prepared by dissolving a non-electrolyte non-volatile solute separately in the solvents X and Y. The molecular weights of the solvents are

$M_X$ and

$M_Y$ , respectively where

$M_X=\displaystyle\frac{3}{4}M_Y$ . The relative lowering of vapour pressure of the solution in X is "m" times that of the solution in Y. Given that the number of moles of solute is very small in comparison to that of solvent, the value of "m" is?

0%
$\displaystyle\frac{3}{4}$
0%
$\displaystyle\frac{1}{2}$
0%
$\displaystyle\frac{1}{4}$
0%
$\displaystyle\frac{4}{3}$

Explanation

THe relationship between molar masses of two solvents is

$\displaystyle M_X=\dfrac {3}{4}M_Y$ ......(1)
THe relative lowering of vapour pressure of two solutions is

$\displaystyle (\dfrac{\Delta P}{P})_X=m(\dfrac{\Delta P}{P})_Y$
But, the relative lowering of vapour pressure of solution is directly proportional to the mole fraction of solute.

$\displaystyle M_x \times \dfrac {5}{1000} = m \times M_Y \times \dfrac {5}{1000}$ .....(2)
Substitute equation (1) in equation (2).

$\displaystyle \dfrac {3}{4} \times M_Y \times \dfrac {5}{1000} = m \times M_Y \times \dfrac {5}{1000}$

$\displaystyle m=\dfrac {3}{4}$
Note:
5 molal solution means 5 moles of solute are dissolved in 1 kg (or 1000 g) of solvent. The number of moles of solvent

$\displaystyle = \dfrac {1000g}{M}$
The mole fraction of solute

$\displaystyle = \dfrac {5}{1000/M}$

$\displaystyle = M \times \dfrac {5}{1000}$

Two open beakers one containing a solvent and the other containing a mixture of that solvent with a non volatile solute are together sealed in a container. Over time:

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The volume of the solution does not change and the volume of the solvent decreases
0%
the volume of the solution deceases and the volume of the solvent increases
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the volume of the solution and the solvent does not change
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the volume of the solution increases and the volume of the solvent decreases

Liquid

$M$ and liquid

$N$ form an ideal solution. The vapour pressures of pure liquids

$M$ and

$N$ are

$450$ and

$700\ mm$

$Hg$ , respectively, at the same temperature. Then correct statement is:
(

${x}_{M}=$ Mole fraction of

$M$ in solution;

${x}_{N}=$ Mole fraction of

$N$ in solution;

${y}_{M}=$ Mole fraction of

$M$ in vapour phase;

${y}_{N}=$ Mole fraction of

$N$ in vapour phase)

0%
$({x}_{M}-{y}_{M})< ({x}_{N}-{y}_{N})$
0%
$\cfrac{{x}_{M}}{{x}_{N}}< \cfrac{{y}_{M}}{{y}_{N}}$
0%
$\cfrac{{x}_{M}}{{x}_{N}}> \cfrac{{y}_{M}}{{y}_{N}}$
0%
$\cfrac{{x}_{M}}{{x}_{N}}= \cfrac{{y}_{M}}{{y}_{N}}$

Explanation

Since

$P^o_N>P^o_M,$ so

$N$ is more volatile than

$M.$

$\therefore y_N>x_N$

and

$x_M>y_M$

Multiply we get

$y_N\times x_M>x_N \times y_M$

$i.e.\ \dfrac{x_M}{x_N}=\dfrac{y_M}{y_N}$

Which one is not equal to zero for an ideal solution?

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$\Delta V_{mix}$
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$\Delta P=P_{observed}-P_{Raoult}$
0%
$\Delta H_{mix}$
0%
$\Delta S_{mix}$

Explanation

For an ideal solution,

(A) The volume of mixing is zero.

$\Delta V_{mix}=0$

(B) The observed pressure is equal to the pressure calculated from Raoult's law, then solution said to be ideal solution.

$P_{observed}=P_{Raoult}$

$\Delta P=P_{observed}-P_{Raoult}=0$

$\Delta H_{mix}=0$ .

(D) But the entropy of mixing is not equal to zero.

$\Delta S_{mix} \neq 0$ .

Option D is correct.

Consider the following liquid - vapour equilibrium Liquid

$\rightleftharpoons$ Vapour. Which of the following relations is correct?

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$\dfrac {dlnG}{dT^{2}} = \dfrac {\Delta H_{v}}{RT^{2}}$
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$\dfrac {dlnP}{dT} = \dfrac {-\Delta H_{v}}{RT}$
0%
$\dfrac {dlnP}{dT^{2}} = \dfrac {-\Delta H_{v}}{T^{2}}$
0%
$\dfrac {dlnP}{dT} = \dfrac {\Delta H_{v}}{RT^{2}}$

Explanation

According to Clausius - Clapeyron's equation

$\dfrac {d \ lnP}{dT}= \dfrac {\Delta H_v}{RT^2}$

According to this equation, the rate at which the natural logarithm of the vapor pressure of a liquid changes with temperature is determined by the molar enthalpy of vaporization of the liquid, the ideal gas constant, and the temperature of the system.

For an ideal solution, the correct option is :

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$\Delta_{mix}S=0$ at constant T and P
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$\Delta_{mix}V\neq 0$ at correct T and P
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$\Delta_{mix}H=0$ at constant T and P
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$\Delta_{mix} G=0$ at constant T and P

Explanation

The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions.

The ideal solutions have two other important properties. The enthalpy of mixing of the pure components to form the solution is zero and the volume of mixing is also zero, i.e.,

For ideal solution,

$\Delta_{mix}H=0$

$\Delta_{mix}V=0$

The van't Hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide is :

0%
2
0%
3
0%
0
0%
1

Explanation

Hint: For strong electrolytes, Van't Hoff factor $(i)$ is equal to the number of ions produced.

Formula:

$\alpha = \dfrac {i-1} {n-1}$

Here,

$\alpha$ is the degree of dissociation,

$i$ van't Hoff factor and

$n$ is number of ions.

Step 1: To find the number of ions in barium hydroxide.

$Ba(OH)_{2}$ is a strong electrolyte that dissociates

$100$ % in an aqueous medium as

$Ba(OH)_{2} (l) \leftrightarrow Ba^{2+} (aq) + 2OH^{-} (aq)$

So,

Number of ions

$n=1+2=3$

Step 2: To find the van't Hoff factor $(i)$ for barium hydroxide.

As barium hydroxide is a strong electrolyte,

$\therefore$

$\alpha=1$

Using formula,

$\alpha = \dfrac {i-1} {n-1}$

$1 = \dfrac {i-1} {3-1}$

$(i-1) = 2$

$i=2+1$

$i=3$

Final answer:

Option

$B$ is the correct answer.

CBSE Questions for Class 12 Engineering Chemistry Solutions Quiz 1 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Solutions Quiz 1 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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