MCQ Questions for Class 12 Engineering Chemistry Solutions Quiz 15

The amount of anhydrous

$Na_{2}CO_{3}$ present in 250 ml of 0.25 M solution is

0%
6.225 g
0%
66.25 g
0%
6.0 g
0%
6.625 g

When NaCl is dissolved in water

0%
boiling point of solution decreases
0%
boiling point of solution increases
0%
boiling point of solution remains unchanged
0%
freezing point of solution increases

Moles of

${\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ to be dissolved in 12 moles of water to lower its vapour pressure by 10mm Hg at a temperature at which vapour pressure of pure water is 50 mm Hg is:

0%
1.5 moles
0%
2 moles
0%
1 mole
0%
3 moles

Vapour pressure of a pure liquid X is 2 atm at 300 K. It is lowered to 1 atm on dissolving 1 g of Y in 20 g of liquid X. If molar mass of X is 200, what is the molar mass of Y?

0%
20
0%
10
0%
100
0%
30

When solute and solvent both are volatile, the vapour pressure of solution is directly proportional to mole fraction of which?

0%
Solute
0%
Solvent
0%
Solute and Solvent
0%
none

Select the correct statement(s) from the following?

0%
Addition of $Hg{I_2}$ to aqueous solution of $KI$ increases the freezing point since a complex ${K_2}\left[ {Hg{I_4}} \right]$ is formed
0%
Three aqueous solutions of $NH_3$ labelled as A, B and C with concentration 0.1 M, 0.01 M and 0.001 M respectively, then the value of Van't Hoff factor for these solution will be in order, ${i_A} < {i_B} < {i_C}$ (consider degree of dilution)
0%
Vapour density of $PC{l_5}\left( g \right)$ dissociating into $PC{l_3}\left( g \right)$ and $C{l_2}\left( g \right)$ is 100, then the Van't Hoff factor will be $4.25$
0%
If $pH$ of $0.1 M$ monobasic acid is found to be 2, its osmotic pressure at a given temperature $T$ kelvin will be $0.11 RT$

Four solutions of

$K_2SO_4$ with the following concentration

$0.1 m$ ,

$0.01 m$ ,

$0.001 m$ and

$0.0001 m$ are available. The maximum value of van't Hoff factor, i, corresponds to:

0%
0.0001 m solution
0%
0.001 m solution
0%
0.01 m solution
0%
0.1 m solution

The vapour pressure of

$CCl_4$ at

$25^0C$ is 143 mm of Hg. 0.5 gm of a non-volatile solute(mol. wt. 65) is dissolved in 100ml of

$CCl_4$ , the vapour pressure of the solution will be:

0%
199.34 mm Hg
0%
143.99 mm Hg
0%
141.43 mm Hg
0%
94.39 mm Hg

$0.6g$ urea is added to

$360g$ water. Calculate lowering in vapor pressure for this solution
(Given: Vapour pressure of

$H_2O$ is

$35mm$ of

$Hg$ )

0%
$0.027mm$ of $Hg$
0%
$0.035mm$ of $Hg$
0%
$0.017mm$ of $Hg$
0%
$0.040mm$ of $Hg$

Explanation

Solution:- (C)

$0.017mm$ of

$Hg$

$\dfrac{P^0-Ps}{P^0}=\dfrac{n}{n+N}$

Lowering in V.P.=

$P^0\times \dfrac{n}{n+N}$

$=35\times \dfrac{\dfrac{0.6}{60}}{\dfrac{0.6}{60}+\dfrac{360}{18}}=0.017mm$ of Hg

In a mixture of

$A$ and

$B$ , having vapour pressure of pure

$A$ and pure

$B$ as

$400mm\, Hg$ and

$600mm\, Hg$ respectively, mole fraction of

$B$ in liquid phases is

$0.5$ . Calculate total vapour pressure and mole fraction of

$A$ and

$B$ in vapour phases.

0%
$500, 0.4, 0.6$
0%
$500, 0.5, 0.5$
0%
$450, 0.4, 0.6$
0%
$450, 0.5, 0.5$

Explanation

$P_T = X_AP_{A^o} + X_BP_{B^o}$

$= 0.5\times 400+0.5\times 600$

$=500nm$ of

$Hg$

$\dfrac{1}{P_T} = \dfrac{y_A}{P^o_A} + \dfrac{1-y_B}{P^o_B}$

$y_A = 0.6$ &

$y_B = 1 -0.6 = 0.4$

When a substance is dissolved in a solvent, the vapour pressure of solvent decrease. It brings

0%
a decrcase in boilings point of solution
0%
an increase in boiling point of the solution
0%
a decrease in freezing point of the solution
0%
an increase in freezeing point of the solution

A sample of pure Cu

$(3.18\,g)$ when heated in the presence of oxygen forms a black oxide of copper

$(CuO)$ . The final weight of CuO is

$3.92$ g. The percent of Cu oxidised is

0%
$65\%$
0%
$69\%$
0%
$76\%$
0%
$98\%$

One mol of a solute A is dissolved in a given volume of a solvent. The association of the solute take place according to

$n A = (A)_n$ . The van't Hoff factor i is expressed as

0%
$i = 1 - x$
0%
$i = 1 + \dfrac{x}{n}$
0%
$i = \dfrac{i - x + \dfrac{x}{n}}{1}$
0%
$i = 1$

How many types of solutions are possible by the combination of any two of three states of matter (solid, liquid, and gas)

0%
$3$
0%
$6$
0%
$9$
0%
$18$

What is the percentage of hydrogen by wt.in vitamin

$C$ ?

0%
$4.55\%$
0%
$41\%$
0%
$20.5\%$
0%
$9.11\%$

Explanation

Mass of

$CO_{2}=85.35-83.85=1.5\ g$
Mass of

$H_{2}O=37.96-37.55=0.41\ g$
Mole of

$CO_{2}\dfrac{1.5}{44}=0.034\ g$
Mass pf

$C=0.034\times 12=0.41\ g$
Mole of

$H_{2}O=\dfrac{0.41}{13}=0.0227$
Mole of

$H=0.0227\times 2=0.0455\ mole$
Mass fo

$H=0.0455\times 1=0.0455\ g$
Mass of oxygen

$=1- (0.41+0.0455)=0.544\ g$

$\% C=\dfrac{0.41}{18}\times 100=41\%$

$\%H =\dfrac{0.0455}{1}\times 100=4.55\%$ Option

$(a)$

What is the percentage of carbon, by wt. in vitamin

$C$ ?

0%
$66.67\%$
0%
` $40.9\%$
0%
$20\%$
0%
$60\%$

Explanation

Mass of

$CO_{2}=85.35-83.85=1.5\ g$
Mass of

$H_{2}O=37.96-37.55=0.41\ g$
Mole of

$CO_{2}\dfrac{1.5}{44}=0.034\ g$
Mass pf

$C=0.034\times 12=0.41\ g$
Mole of

$H_{2}O=\dfrac{0.41}{13}=0.0227$
Mole of

$H=0.0227\times 2=0.0455\ mole$
Mass fo

$H=0.0455\times 1=0.0455\ g$
Mass of oxygen

$=1- (0.41+0.0455)=0.544\ g$

$\% C=\dfrac{0.41}{18}\times 100=41\%$ Option

$(b)$

1.0 g of moist sample of a mixture of

$KCl$ and

$KClO_3$ was dissolved in water and made up to 250mL. 25 mL of this solution was treated with

$SO_2$ . The chlorate was reduced to chloride and an excess of

$SO_2$ was removed by boiling. The total chloride was precipitated as

$AgCl$ weighing 0.1435 g. In another experiment, 25 mL of the original solution was heated with 30 mL of 0.2 N solution of ferrous sulphate and the unreacted ferrous sulphate required 37.5 mL of 0.08 N solution of an oxidising agent for complete oxidation. For the given reaction:

$ClO_3^{ɵ} +6Fe^{2+}+6H^{+} \rightarrow Cl^{ɵ}+6Fe^{3+}+3H_2O$

What is the mass per cent of potassium chloride in the moist sample?

0%
37.25%
0%
61.25%
0%
3.725%
0%
74.5%

What is the percentage of silica present in the ore, by weight?

0%
$2.82$
0%
$3.02$
0%
$0.465$
0%
$6.05$

Explanation

Mole wt of kaolin

$=2589/mole$
Mole of

$Al$ is kaolin

$=\dfrac{13}{258}=0=5$

$(Al_{2}O_{3})$
Mole wt of Gibbrite

$=156 9/mole$
Mole of gibbrite

$=\dfrac{87}{156}=0.557\ mole$
Total mole of

$Al_{2}O_{3}=0.557+0.05=0.607\ mole$
Mole of

$HO_{2}$ from kaolin

$=2\times 0.05=0.01\ mole$
Max mole of Muld forms

$=\dfrac{0.1}{5}=0.02\ mole$

$1$ mole Mud

$=3$ mole

$N_{2}O_{3}$

$0.02$ mole Mud

$=x-1$

$x=3\times 0.02=0.06$
Max

$Al_{2}$ can be expected

$=0.607-0.06=0.547$

$\% Al$ can be expected

$=\dfrac{0.547}{0.607}\times 100=90.11\%$

mole

$SiO_{2}=0.1\ mole$
Max

$SiO_{2}=0.1\times 60=6g$

$\%$ sodia

$=\dfrac{6}{15}\times 100=6\%$ Option

$(d)$

Calculate van't Hoff factor for

$0.2\ m$ aqueous solution of

$KCl$ which freezes at

$-0.680^o$ C. (

$K_f=1.86$ K kg

$mol^{-1}$ )

0%
$3.72$
0%
$1.83$
0%
$6.8$
0%
$1.86$

Explanation

Given,

$m =0.2m$

$K_f=1.86\ K\ Kg\ mol^{-1}$

$T_f=-0.680^0C$

We have,

$(\Delta T_f)$

$=i\times K_f \times m$

Here,

$i=$ Van't Hoff factor

$\Delta T_f=0-0.680=0.680^0C$

Here,

$i=\dfrac{\Delta T_f}{K_f\times m}=\dfrac{0.68}{1.86\times 0.2}=1.83$

Hence, option

$B$ is correct.

$1.50\ g$ sample of type metal (an alloy of

$Sn, pb, Cu$ and

$Sb$ ) is dissolved in nitric acis and metastannic acid,

$H_2SnO_3$ , prpitates. This is delydrated by heating to tin

$(IV)$ oxide, which is found to weight

$0.50\ g$ . What percentage of tin was in the original type metal sample?

$(Sn=119)$

0%
$33.33\%$
0%
$26.27\%$
0%
$29.38\%$
0%
$52.54\%$

Explanation

weight of

$SnO_2=0.50\ g$

$151\ g\ SnO_2=119\ g\ Sn$

$0.50\ SnO_2=\dfrac {119\times 0.50}{151}$

$=0.3940\ g\ Sn$

$\%$ of

$Sn=\dfrac {0.3940}{1.50}\times 100$

$=26.269\%$

$\%$ of

$Sn=26.27\%$

When the air temperature is below freezing, the saturation vapor pressure over water is _______.

0%
Equal to zero
0%
Less than the saturation vapor pressure over ice
0%
Greater than the saturation vapor pressure over ice
0%
Equal to the saturation vapor pressure over ice

To determine soluble (free)

$SiO_2$ in a rock, an alkaline extraction was carried out, as a result of which there was found

$1.52\%$ of

$SiO_2$ in the extract and also

$1.02\%$ of

$Al_2O_3$ . Considering that, apart from the free

$SiO_2$ , the extract also contained the

$SiO_2$ , that had passed into it from Kaolin

$(2SiO_2, Al_2O_3)$ , the percentage of

$SiO_2$ , in the rock being analysed is

$(Si=28, Al=27)$

0%
$1.20$
0%
$0.32$
0%
$0.50$
0%
$1.52$

Explanation

Total

$\%$ of

$SiO_2$ observed

$=1.52\%$

$\%$ of

$Al_2O_3=1.02\%$

$\%$ of

$SiO_2=$ out of

$SiO_2$ from

$2SiO_2.Al_2O_3$
And

$Al_2O_3$ is from only

$2SiO_2.Al_2O_3$
Mass of

$SiO_2$ in

$2SiO_2.Al_2O_3=2\times (38+32)$

$=120\ gm$
Mass of

$Al_2O_3$ in

$2SiO_2.Al_2O_3 =27\times 2+16\times 3$

$=102\ gm$
Now

$102\ gm\ Al_2O_3$ contribute

$102\%$

$\therefore \ 1\ gm$ contribute

$0.01\%=\dfrac {1.02}{102}\%$

$\therefore \ 120\ gm\ SiO_2.Al_2O_3$ contribute

$120\times 0.01=1.20\%$

$\therefore \ \%$ of

$SiO_2$ in nock

$=1.52-1.20$

$=0.32\%$

How much slurry obtained per hour?

0%
$5000\ kg$
0%
$500\ kg$
0%
$19,500\ kg$
0%
$20,000\ kg$

Explanation

$2\%NaOH=\dfrac{x}{25000}\times 100$

$x=\dfrac{2\times 25000}{100}=500\ kg\ NaOH$ & water

$18\%=NaCl=\dfrac{x}{25000}\times 100$

$x=\dfrac{18\times 25000}{100}=4500\ kg\ NaCl$

$90\% NaCl=\dfrac{4500}{y}\times 100$

$y=\dfrac{4500}{90}\times 100=5000$

$5\%=$ water on slurry & Care soln

$=500$
man of mole in feed streak

$=0.8\times 2.500=20000$
water evaporated

$=20,000-500=19,500$
Concentrated soln

$=x=$ max of

$NaOH+$ water

$=500\ kg$
Mass of slurry

$=y=5000\ kg$ Option

$(a)$

How much water is evaporated per hour?

0%
$500\ kg$
0%
$500\ kg$
0%
$19,500\ kg$
0%
$20, 000\ kg$

Explanation

$2\%NaOH=\dfrac{x}{25000}\times 100$

$x=\dfrac{2\times 25000}{100}=500\ kg\ NaOH$ & water

$18\%=NaCl=\dfrac{x}{25000}\times 100$

$x=\dfrac{18\times 25000}{100}=4500\ kg\ NaCl$

$90\% NaCl=\dfrac{4500}{y}\times 100$

$y=\dfrac{4500}{90}\times 100=5000$

$5\%=$ water on slurry & Care soln

$=500$
man of mole in feed streak

$=0.8\times 2.500=20000$
water evaporated

$=20,000-500=19,500$ Option

$(c)$

How much concentrated solution obtained per hour?

0%
$5000\ kg$
0%
$500\ kg$
0%
$19,500\ kg$
0%
$20,000\ kg$

Explanation

$2\%NaOH=\dfrac{x}{25000}\times 100$

$x=\dfrac{2\times 25000}{100}=500\ kg\ NaOH$ & water

$18\%=NaCl=\dfrac{x}{25000}\times 100$

$x=\dfrac{18\times 25000}{100}=4500\ kg\ NaCl$

$90\% NaCl=\dfrac{4500}{y}\times 100$

$y=\dfrac{4500}{90}\times 100=5000$

$5\%=$ water on slurry & Care soln

$=500$
man of mole in feed streak

$=0.8\times 2.500=20000$
water evaporated

$=20,000-500=19,500$
Concentrated soln

$=x=$ max of

$NaOH+$ water

$=500\ kg$ Option

$(b)$

$200\ g$ of a sample of olem labelled as

$109.0\% ,\ 12\ g$ water is added. The new labelling of the oleum sample is

0%
$106.0\%$
0%
$103.0\%$
0%
$102.8\%$
0%
$105.6\%$

Explanation

Given:

$200\ gm$ of sample of oleum
labelled

$109\%$

$12\ gm\ H_2O$ is added

$503+H_2O\to H_2SO_4$

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

$109\%$ means

$100\ gm$ sample of okeam

$(SO_3+H_2SO_4)$ in which

$9\ gm\ H_2O$ react with

$\% SO_3$

$18\ gm\ H_2O\to 80\ gm\ SO_3$

$\therefore \ 9\ gm\ H_2O\to \dfrac {9}{18}\times 80=40\ gm\ SO_3$
in

$100\ gm$ sample

$\therefore \ wt$ of

$H_2SO_4=60\ gm$ in

$100\ gm$ sample
Now,

$12\ gm\ H_2O$ is added

$18\ gm\ H_2O\to 80\ gm\ SO_3$

$\therefore \ 12\ g\ H_2O\to \dfrac {12}{18}\times 80$

$=53.33\ gm\ SO_3$ reacted to from

$(53.33+12)\ gm\ H_2SO_4=65.33\ gm\ H_2SO_4$
Now,
In

$200\ g$ sample :-

$SO_3=80\ gm$

$H_2SO_4=120\ gm$
After adding

$H_2O; \ SO_3=80--53.33\ gm =26.67\ gm$

$H_2SO_4=120+65.33\ gm$

$=185.33\ gm$

$\%$ of

$SO_3=\dfrac {26.67}{212}\times 100=12.67$

$\therefore \ 12.6\ gm\ SO_3$ react with:-

$2.8\ gm\ H_2O$

$\therefore \ \%$ label on sample

$=100+2.8=102.8\%$

Directions: In the following questions, a statement of assertion is followed by a statement of reason. Mark the correct choice:

Assertion: Unseeing of vapour pressure is not dependent on the number of species present in the solution.
Reason : Lowering of vapour pressure and relative lowering of vapour pressure are colligative properties.

0%
If both assertion and reason are true and reason is the correct explanation of assertion.
0%
If both assertion and reason are true but reason is not the correct explanation of assertion.
0%
If assertion is true but reason is false.
0%
If both assertion and reason are false.

Directions: In the following questions, a statement of assertion is followed by a statement of reason. Mark the correct choice:

Assertion: The vapour pressure of an aqueous solution is less than 1.013 bar at 373.15 K.
Reason: Vapour pressure of water is 1.013 bar at 373.15K.

0%
If both assertion and reason are true and reason is the correct explanation of assertion.
0%
If both assertion and reason are true but reason is not the correct explanation of assertion
0%
If assertion is true but reason is false.
0%
If both assertion and reason are false.

Which of the following is correct for an ideal solution?

0%
$\Delta H_{mix} = 0$
0%
$\Delta S_{mix} = 0$
0%
$\Delta G_{mix} = 0$
0%
None of these

$2.0\ g$ mixture of sodium carbonate and sodium bicarbonate on heating to constant weight gave

$224\ mL$ of

$CO_{2}$ at N.T.P.
The

$\%$ weight of sodium bicarbonate in the mixture is:

0%
$50$
0%
$54$
0%
$80$
0%
$84$

Explanation

On heating the mixture only sodium bicarbonate dissociates. Sodium carbonate does not dissociate due to its stability.

$2NaHCO_3 + heat \rightarrow Na_2CO_3 + CO_2 + H_2O$

Let the weight of

$Na_2CO_3$ in the mixture be

$x$ g

Weight of

$NaHCO_3$ in the mixture

$=(2-x)$ g

No. of moles of

$Na_2CO_3$

$=\dfrac {x}{106}$ moles

No. of moles of

$NaHCO_3$

$=\dfrac{(2-x)}{84}$ moles

From the equation:

2 moles of

$NaHCO_3$ give 1 mole of

$CO_2$ gas

$\dfrac{(2-x)}{84}$ moles of

$NaHCO_3$ give

$\dfrac{(2-x)}{84} \times \dfrac {1}{2} = \dfrac {(2-x)}{168}$ moles of

$CO_2$ gas

At N.T.P

22400 ml of

$CO_2$ gas = 1 mole

224 ml of

$CO_2$ gas =

$\dfrac{1}{22400} \times 224 = 0.01$ moles

Now

$\dfrac {(2-x)}{168} = \dfrac {1}{100}$

$\Rightarrow x = 0.32$ g

Weight of sodium bicarbonate

$=(2-0.32) g = 1.68$ g

Weight percentage of sodium bicarbonate =

$\dfrac{1.68}{2} \times 100 = 84$ %

The compound

$Na_2 CO_3 . x H_2O$ has

$50 \% H_2O$ by mass. The value of "x" is :

0%
4
0%
5
0%
6
0%
7
0%
8

Explanation

(i) Molar mass of

$Na_2CO_3 = 106$ unit

Since mass % of the water in the compound is 50%.

$\Rightarrow$ Mass of the hydrated compound

$= 2\times 106=212g$

i.e. Mass of water

$=106 u$ per hydrated compound.

$\therefore$ Number of water molecules,

$x=\dfrac{mass}{molecular \ mass}=\dfrac{106}{18}\approx 6$

Hence option C correct.

The vapour pressure of pure liquid A at

$27^0 C$ is

0%
60 torr
0%
48 torr
0%
49.26 torr
0%
61.58 torr

If the experimenty is performed aqueous

$AlCl_3$ solution (a= 0.8 ) prepared by dissolving 1 mole of

$AlCl_3$ in 17 mole of water and the decrease in the mass of solution in the experiment is found to be 0.18 g then the increase in the mass of absorbent should be

0%
0.216 g
0%
0.225 g
0%
0.191 g
0%
1.08 g

Explanation

$AlCl_3 \rightleftharpoons Al^{3+} +3Cl^{-}$

$1-0.8\\=0.2$

$0.8$

$3\times 0.8=\\2.4$

Total no. of mole

$=0.2 + 0.8 +2.4 =3.4$

Now, decrease in mass of solution

$\propto P$
and increase in mass of absorber

$\propto P^o$

$\therefore \cfrac{P^o}{P}=X_2=\cfrac{17}{17+3.4}=\cfrac{5}{6}=\cfrac{0.18}{|\Delta m|_{absorber}}$

$\therefore$ Increase in mass of absorber

$=0.216\ gm$

$80^o C$ , the vapour pressure of pure liquid A is 520 mm Hg and that of pure liquid B is 1000 mm Hg. If a mixture solution of A and B boils at

$80^o C$ and 1 atm pressure, the amount of A in the mixture is :

(1 atm=760 mm Hg)

0%
34 mol%
0%
48 mol%
0%
50 mol%
0%
52mol%

Sugar in urine sample can be detected by:

0%
Fehling's solution
0%
Benedict's solution
0%
Tollen's reagent
0%
all the three $(a), (b)$ and $(c)$

$50 mL$ of solution of

$Na_2CO_3$ neutralizes

$49.35 mL$ of

$4 N \ HCl$ . the reaction is represented as

$CO^{2-}_3 +2H^+ \rightarrow CO_2 +H_2O$
The density of this

$Na_2CO_3$ solution is 1.25 g

$mL^{-1} ,$ the percentage of

$Na_2CO_3$ in it is :

0%
$47.7$
0%
$37.7$
0%
$26.7$
0%
$16.7$

Dry air is bubbled successively through (i) a solution, (ii) its solvent, and (iii) through

$CaCl_2$ . The lowering of vapour pressure of the solvent due to the addition of solute is equal to

0%
$p^0 \times \dfrac {loss\ in\ wt. of\ solvent}{gain\ in\ wt\ of\ CaCl_2}$
0%
$\dfrac {loss\ in\ wt. of\ solvent}{gain\ in\ wt\ of\ CaCl_2}$
0%
$\dfrac {loss\ in\ wt. of\ solution}{gain\ in\ wt\ of\ CaCl_2}$
0%
$\dfrac {loss\ in\ wt. of\ solute}{gain\ in\ wt\ of\ CaCl_2}$

The reading of pressure gauge at bubble point is :

0%
500
0%
600
0%
700
0%
None

Explanation

$X_A = 0.75 \quad X_B = 0.05$ above 500 mm Hg only liquid phase exists .

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

$P_{cqycqfyr fcUnq} = X_A P60_A + X_B P^0_B$

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

$y_A = 0.75 \quad \quad y_B = 0.5$
at dew point

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

$\Rightarrow P_T = \frac {800}{1.75} = 457.14 mm Hg$
Below dew point only vapor phase exists.

The reading of pressure gauge at which only vapour phase exists is

0%
501
0%
457.14
0%
425
0%
525

Explanation

$X_A = 0.75 \quad X_B = 0.05$ above 500 mm Hg only liquid phase exists .

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

$P_{cqycqfyr fcUnq} = X_A P60_A + X_B P^0_B$

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

$y_A = 0.75 \quad \quad y_B = 0.5$
at dew point

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

$\Rightarrow P_T = \frac {800}{1.75} = 457.14 mm Hg$
Below dew point only vapor phase exists.

The vapour pressure of two micible liquid A and B are 300 and 500 mm of Hf respectively. in a flask , 2 moles of A are mixed with moles of B. further to the mixture , 32 g of an ionic non -volatile solute MCI (partially ionised, mol mass = 70 u) were also added. thus, the final vapour pressure of solution was found to be 420 mm and Hg. Then, identify the correct statement(s). (Assume the liquid mixture of A and B to behave ideally).

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The numerical value of relative lowering in vapour pressure upon addition of solute MCI is 1/15
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Th solute MCI is 25 % ionised in the above question.
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The solute MCI is 23.33 % ionised in the above question
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Upon addition of excess $Pb(NO_3)_2$ the number of moles of $PbCl_2$ precipitated is 2/ 35 .

Explanation

$P_0= X_A P^0_A + X_B P^0_B = \frac {2}{8} \times 300 + \frac {6}{7} \times 500 = 450$ mm of Hg
Now RLVP =

$\frac {P_0 -P_s}{P_0} = \frac { 450 - 420}{450} = \frac {1}{15}$
Also RLVP =

$( \frac {in}{In +N } )$

$\frac {1}{15} = \frac { \frac { 32i}{70} }{ \frac {32i}{70} + 8 }$

$\therefore i - 1.25 = ( 1+ ( 2-1 ) \alpha ).$ .
So

$\alpha$ = 0.25 ( or 25 %)

$\therefore n_{CI^- }$ produced =

$\frac {25}{100} \times \frac {32}{70} = \frac {4}{35}$

$\therefore n_{PbCl_2 }$ precipitated =

$\frac {1}{2} \times \frac {4}{35} = \frac {2}{35}$

CBSE Questions for Class 12 Engineering Chemistry Solutions Quiz 15 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Solutions Quiz 15 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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