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CBSE Questions for Class 12 Engineering Chemistry Solutions Quiz 5 - MCQExams.com
CBSE
Class 12 Engineering Chemistry
Solutions
Quiz 5
At $$25^oC$$, the total pressure of an ideal solution obtained by mixing 3 mole of A and 2 mole of B, is 184 torr. What is the vapour pressure (in torr) of pure B at the same temperature (Vapour pressure of pure A at $$25^oC$$ is 200 torr.)?
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0%
180
0%
160
0%
16
0%
100
Explanation
$$x_A=\dfrac{n_A}{n_A+n_B}=\dfrac{3}{3+2}=\dfrac{3}{5}=0.6$$
$$x_B=0.4$$
$$p=p^0_A+p^0_Bx_B$$
$$184=200\times 0.6+p^0_B\times 0.4$$
$$184=120+p^0_B\times 0.4$$
$$p^0_B=\dfrac{64}{0.4}=160 torr$$
An aqueous solution containing $$28\%$$ by mass of a liquid A (molecular mass = 140) has a vapour pressure of 160 mm at $$37^oC$$. Find the vapour pressure of pure liquid A (the vapour pressure of water $$37^oC$$ is 150 mm).
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$$p_A=358.3 mm$$
0%
$$p_A=388.3 mm$$
0%
$$p_A=258.3 mm$$
0%
None of these
Among the following, that does not form an ideal solution is:
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$$C_6H_6$$ and $$C_6H_5CH_3$$
0%
$$C_2H_5Cl$$ and $$C_2H_5OH$$
0%
$$C_6H_5Cl$$ and $$C_6H_5Br$$
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$$C_2H_5Br$$ and $$C_2H_5I$$
The depressions in freezing point for 1 m urea, 1 m glucose and 1 m NaCl are in the ratio:
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1 : 2 : 3
0%
3 : 2 : 2
0%
1 : 1 : 2
0%
None of these
Explanation
We have,
$$\Delta T_f = i K_f m$$
$$i_{urea}= i_{glucose}=1, i_{NaCl}=2$$
So, we have ratio for $$\Delta T_f$$-
$$1\times K_f\times 1 : 1\times K_f\times 1 : 2\times K_f\times 1=1:1:2$$
Hence, option (C) is correct.
Which one of the following is a primary standard?
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Oxalic acid
0%
Sodium thiosulphate
0%
Sodium hydroxide
0%
Potassium dichromate
Explanation
The primary standard is a compound of sufficient purity from which standard solutions of known normalities can be prepared by direct weighing of it and diluting to a defined volume of solution.
Potassium Dichromate$$(K_2Cr_2O_7)$$ is suitable to be used as a primary standard. It cannot be obtained in very pure form. It readily reacts with any traces of organic material or any other reducing substance in water.
18 g glucose is dissolved in 90 g of water. The relative lowering of vapour pressure of the solution is equal to__________.
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0%
6
0%
0.2
0%
5.1
0%
0.02
Explanation
Molecular mass of water$$=$$ $$2 \times1+1 \times 16=18g$$
=2×1+1×16=18g
For $$90g$$
178.2g
water $$n_A=5$$
Molecular mass of glucose $$6 \times 12+12 \times 1+6 \times16=180g$$
For $$18g$$ glucose $$n_B=0.1$$
Now, $$X_B=\dfrac {0.1}{0.1+5}=0.0196$$
~$$0.02$$
X is a non-volatile solute and Y is volatile solvent. The following vapour pressures are obtained by dissolving X in Y.
X/mol $$L^{-1}$$
Y/mm Hg
0.1
$$P_1$$
0.25
$$P_2$$
0.01
$$P_3$$
The correct order of vapour pressure is :
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$$P_1 < P_2 < P_3$$
0%
$$P_3 < P_2 < P_1$$
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$$P_3 < P_1 < P_2$$
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$$P_2 < P_1 < P_3$$
Explanation
$$\begin{array}{l}\text { Lesser is the concentration of solution greater is the vapour pressure }\\ \\\qquad \begin{array}{c} P_{2}<P_{1}<P_{3} \\ (0.25 M)(0.1 M)(0.01 M)\end{array} \\\\\text {Hence, option B is correct.}\end{array}$$
The freezing point depression constant for water is $$1.86^oC$$ $$m^{-1}$$. If $$5$$ g $$Na_2SO_4$$ is dissolved in 45 g $$H_2O$$, the freezing point is changed by $$-3.82^oC$$.
Calculate the van't Hoff factor for $$Na_2SO_4$$.
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$$0.381$$
0%
$$2.05$$
0%
$$2.63$$
0%
$$3.11$$
Explanation
Given: $$K_f =1.86^oC m^{-1}$$, $$w_B= 5g,w_A= 45g, M_B= 142 g/mol$$
$$\triangle T=i\times K_f\times \dfrac{w_B\times 1000}{M_B\times w_A}$$
$$3.82=i\times 1.86\times \dfrac{5\times 1000}{142\times 45}$$
$$i=2.63$$
An aqueous solution is $$1$$ molal in $$KI$$. Which change will cause the vapour pressure of the solution to increase?
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Addition of $$NaCl$$
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Addition of $$Na_2SO_4$$
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Addition of $$1$$ molal $$KI$$
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Addition of water
If the various terms in the following expressions have usual meanings, the van't Hoff factor 'i' cannot be calculated by which of the following expression?
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$$\pi V=\sqrt{i}nRT$$
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$$\triangle T_f=i\times K_f\times m$$
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$$\triangle T_b=i\times K_b\times m$$
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$$\dfrac{p^o_{solvent}-p_{solution}}{p^o_{solvent}}=i\begin{bmatrix}\dfrac{n}{N+n}\end{bmatrix}$$
Which of the following property indicates weak intermolecular forces of attraction in liquid?
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High heat of vaporization
0%
High vapour pressure
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High critical temperature
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High boiling point
Explanation
The molecules that are having weak intermolecular forces of attraction, they can easily dissociate and convert into vapour state,
so, molecules with weak intractions have high vapour pressure.
Hence, option $$(B)$$ is correct.
Choose the correct statement, when the concentration of a salt solution is increased:
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Boiling point increases while vapour pressure decreases
0%
Boiling point decreases while vapour pressure increases
0%
Freezing point decreases while vapour pressure increases
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Freezing point increases while vapour pressure decreases
Explanation
The
vapour
pressure depends upon the escape of solvent molecules from the surface of the liquid. When
concentration
of
salt
solution is increased,
lesser
number of solvent molecules will escape into
vapours
. It means,
vapour
pressure of the solution will be less than that of the pure solvent and the boiling point of the solution is inversely proportional to the
vapour
pressure.
Boiling point $$\propto \dfrac { 1 }{ Vapour\ pressure } $$
That is why volatile solvent having a very low B.P. possesses a high vapour pressure.
Two liquids $$A$$ and $$B$$ form an ideal solution. At $$300K,$$ the VP of solution containing one mole of $$ 'A'$$ and $$4$$ mole $$'B'$$ is $$560 mm $$ $$Hg. $$ At the same temp, if one mole of $$'B'$$ is taken out from the solution the VP of the solution has decreased by $$10mm$$ $$Hg,$$ the VP of pure $$A\ \& \ B$$ are (in mm):
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$$ 400, 600$$
0%
$$500,500$$
0%
$$300, 700$$
0%
$$200,800$$
If in a solvent, n simple molecules of solute combine to form an associated molecule, $$\alpha$$ is the degree of association, the van't Hoff's factor is equal to :
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$$\cfrac {1}{1-n\alpha }$$
0%
$$1 -\alpha +n\alpha $$
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$$1-\alpha +\cfrac {\alpha}{n}$$
0%
$$\cfrac {\alpha}{n} -1 +\alpha$$
Explanation
$$nA \ \ \ \ \rightleftharpoons \ \ \ \ An$$
$$1- \alpha$$ $$\dfrac {\alpha}{n}$$
$$i = 1- \alpha + \dfrac{\alpha}{n}$$
Hence, option C is correct.
Van't Hoff factor, when benzoic acid is dissolved in benzene, will be:
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$$2$$
0%
$$1$$
0%
$$0.5$$
0%
$$1.5$$
Explanation
Benzoic acid dissolved in benzene is found to have molecular mass double of its normal molecular mass. This is explained by the fact that benzoic acid forms a dimer in benzene solution due to hydrogen bonding.
$$ \displaystyle i= \dfrac { \text { n (observed) } }{ \text { n (theoretical) } }$$
$$ \displaystyle i=\dfrac { \text { 1 } }{ \text { 2 } }$$
$$ \displaystyle i=0.5$$
$$ \displaystyle \text {n (observed)}=$$ number of solute particles present in solution.
$$ \displaystyle \text { n (theoretical)}=$$ number of solute particles without considering association/dissociation.
Van't Hoff factor, when benzoic acid is dissolved in benzene, will be $$0.5$$.
Hence, the correct answer is $$\text{option C}$$.
The vapour pressure of a solvent decreased by $$10\ mm$$ of mercury when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is $$0.2$$. What should be the mole fraction of the solvent, if the decrease in the vapour pressure is to be $$20\ mm$$ of mercury?
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$$0.8$$
0%
$$0.6$$
0%
$$0.4$$
0%
$$0.2$$
Explanation
$$\text{Mole fraction of solute}=\cfrac { \text{lowering in vapour pressure} }{\text{ vapour pressure of solvent} } $$
comparing under the two conditions
$$\cfrac { 0.2 }{ \text{mole fraction of solute} } =\cfrac { 10 }{ 20 } \quad \quad $$
or, the mole fraction of solute $$=0.4$$
mole fraction of solvent $$=(1-0.4)=0.6$$
Hence, the correct option is $$\text{B}$$
At a constant temperature, which of the following aqueous solutions will have the maximum vapour pressure?
[Mol weight: $$NaCl=58.5,{ H }_{ 2 }{ SO }_{ 4 }=98.0g.{ mol }^{ -1 }$$]
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$$1$$ molal $$NaCl$$
0%
$$1$$ molar $$NaCl$$
0%
$$1$$ molal $${H}_{2}{SO}_{4}$$
0%
$$1$$ molar $${H}_{2}{SO}_{4}$$
Explanation
One molar (1M) aqueous solution is more concentrated than one molal aqueous solution of the same solute.
In solution, $${H}_{2}{SO}_{4}$$ provides three ions, while $$NaCl$$ provides two ions. Hence, the
vapour
pressure of a
solution
of $$NaCl$$ is higher (as it gives
less number of
ions).
Therefore, 1 molal $$NaCl$$ will have the maximum
vapour
pressure.
Hence, option $$A$$ is correct.
Two liquids $$A$$ and $$B$$ form an ideal solution. What is the vapour pressure of solution containing $$2$$ moles of $$A$$ and $$3$$ moles of $$B$$ at $$300\ K$$?
[Given: At $$300\ K$$, vapour pressure of pure liquid $$A(P_{A}^{\circ}) = 100\ torr$$, Vapour pressure of pure liquid $$B(P_{B}^{\circ}) = 300\ torr]$$.
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$$200\ torr$$
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$$220\ torr$$
0%
$$180\ torr$$
0%
None of these
Explanation
$$x_{A}=\cfrac {2}{2+3}=\cfrac{2}{5}, x_B=\cfrac {3}{5}$$
Vapour pressure of solution
$$=x_AP^o_A+x_BP^o_B\quad$$ [Raoult's Law]
$$=\cfrac {2}{5}\times 100+\cfrac {3}{5}\times 300$$
$$=220$$ $$torr$$
A cylinder filled with a movable piston contains liquid water in equilibrium with water vapour at $${ 25 }^{ o }C$$. Which one of the following operations results in a decrease in the equilibrium vapour pressure?
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Moving piston downward a short distance
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Removing a small amount of vapour
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Removing a small amount of the liquid water
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Dissolving salt in the water
Explanation
Vapour pressure is a surface phenomenon, when the surface area is more, vapour pressure will be more. When we add solute, what we do is we decrease the vapour pressure of solvent because same of the salt ions will be present at the solvent surface, thus taking place of same of the solvent particles which were actively participating in maintain, that previous high value of vapour pressure. So, the equilibrium vapour pressure value decreases. Thus, dissolving salt in water results in decreasing of equilibrium vapour pressure.
On the surface of the earth at $$1\ atm$$ pressure, a balloon filled with $$H_{2}$$ gas occupies $$500\ mL$$. This volume is $$5/6$$ of its maximum capacity. The balloon is left in air. It starts rising. The height above which the balloon will burst if temperature of the atmosphere remains constant and the pressure decreases $$1\ mm$$ for every $$100\ cm$$ rise of height is:
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$$120\ m$$
0%
$$136.67\ m$$
0%
$$126\ m$$
0%
$$100\ m$$
Explanation
Given:
$$\ P_1 = 1 atm = 760 mm of Hg $$
$$\ V_1 = 500 ml $$
Solution:
Let Maximum capacity $$ \ V_2 $$= x
According to question ,
$$ 500 = \dfrac {5}{6} \times x $$
$$\Rightarrow x = 500 × \dfrac{6}{5} $$
$$\Rightarrow x = 600 $$
From the ideal gas law ,
$$\ P_1V_1 = P_2V_2 $$
$$\Rightarrow 760 × 500 = \ P_2 × 600 $$
$$\Rightarrow \ P_2 = 760 × \dfrac{500}{600} $$
$$\Rightarrow \ P_2 = 633.33 mm of Hg $$
Change in pressure = $$\ P_2 - \ P_1 $$
$$\Rightarrow Change\ in\ pressure\ = 760 - 633.33 $$
$$\Rightarrow Change\ in\ pressure\ = 126.67 mm of Hg $$
It is given that for each 1 mm decrease, rise = 100 cm
Therefore, Height = 100 × 126.67 cm
$$\Rightarrow$$ Height = 126.67 m
Hence, option C is correct and the answer is 126.67m.
Lowering of vapour pressure in $$2$$ molal aqueous solution at $$373K$$ is____________.
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$$0.35$$ bar
0%
$$0.022$$ bar
0%
$$0.22$$ bar
0%
$$0.035$$ bar
Explanation
Given
A solute B have 2 moles in 1 kg of water
Solution
Molality=1
No.of moles of B=Xb
Molality=Xb*1000/(1-Xb)*18
Putting values
Xb=0.0347
No.of moles of Water=Xa
Xa=1-Xb
Xa=0.9653
Pressure of 1 mole =760mmHg
Pressure of 0.9653 mol=760*0.9653=733.628mmHg
Lowering of Vapour pressure=760-733.628=26.372mmHg=0.035 bar
The correct option is D
The vapour pressure of pure liquid $$A$$ at $$300\ K$$ is $$577\ Torr$$ and that of pure liquid $$B$$ is $$390\ Torr$$. These two compounds form ideal liquid and gaseous mixtures. Consider the equilibrium composition of a mixture in which the mole fraction of $$A$$ in the vapour is $$0.35$$. Find the mole % of $$A$$ in liquid.
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0%
0.628
0%
0.872
0%
0.267
0%
0.834
Explanation
$$P^o_A= 577 torr, P^o_B= 390torr, {x_A}'=0.35$$
Mole fraction of $$A$$ in vapour phase= $$\cfrac {P_A}{P}$$
$$P_A=x_AP^o_A, P=P_A+P_B=x_AP^o_A+(1-x_A)P^o_B$$
$$\therefore 0.35= \cfrac {x_A\times 577}{x_A\times 577+(1-x_A)390}$$
$$\therefore x_A= 0.267$$
The freezing point depression constant for water is $$-1.86C$$. If $$5.00$$ g of $${ Na }_{ 2 }{ SO }_{ 4 }$$ is dissolved in $$45.0$$ g $${ H }_{ 2 }O$$. The freezing point is changed by $$-3.82C$$.
Calculate the van't Hoff factor?
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$$2.63$$
0%
$$0.311$$
0%
$$0.38$$
0%
$$3.8$$
Explanation
Depression in freezing point= $$\Delta T_f=iK_fm$$
Given values, $$\Delta T_f=-3.82^oC$$
$$K_f=-1.86^oC$$
$$=?$$
Molality= $$\cfrac {5}{142}\times \cfrac {1000}{45}=0.7824m$$
$$\Delta T_f=iK_fm$$
$$\Rightarrow i=\cfrac {\Delta T_f}{K_fm}=\cfrac {-3.82}{-1.86\times 0.7824}$$
$$\Rightarrow i=\cfrac {3.82}{1.455}=2.62$$
For the reversible reaction.
$$N_2 (g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$$
at $$500^o$$ C, the value of $$K_p$$ is $$1.44 \times 10^{-5} \ atm^{-2}$$ when partial pressure is measured in atmosphere. The corresponding value of $$K_c$$, with concentration in $$mol/ L$$ is:
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$$\displaystyle \frac{1.44 \times 10^{-5}}{(0.082 \times 500)^{-2}}$$
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$$\displaystyle \frac{1.44 \times 10^{-5}}{(8.314 \times 773)^{-2}}$$
0%
$$\displaystyle \frac{1.44 \times 10^{-5}}{(0.082 \times 773)^{2}}$$
0%
$$\displaystyle \frac{1.44 \times 10^{-5}}{(0.082 \times 773)^{-2}}$$
Explanation
For any given equilibrium, $$K_P=K_C(RT)^{\Delta ng}$$
where, $$\Delta ng$$= (gaseous moles of products)-(gaseous moles of reactants)
In given reaction, $$\Delta ng= 2-(3+1)=-2$$
$$\therefore K_P=\cfrac {K_C}{(RT)^2}$$
$$\therefore K_C=K_P(RT)^2=\cfrac {K_P}{(RT)^{-2}}=\cfrac {1.44 \times 10^{-5}}{(0.082 \times 773)^{-2}}mol/L$$
If relative decrease in vapour pressure is $$0.4$$ for a solution containing $$1$$ mol $$NaCl$$ in $$3$$ mol $$H_{2}O$$.
$$NaCl$$ is_____ % ionized.
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$$60$$%
0%
$$50$$%
0%
$$100$$%
0%
$$40$$%
Explanation
no. of moles of $$NaCl$$=1
relative lowering of vapour pressure =$$i \times {\text{mole fraction of solute}}$$
relative lowering of vapour pressure= $$ \dfrac {i \times {n(solute)}}{n(solute) + n(solvent)}$$
0.4=$$i \times \dfrac{1}{4}$$
therefore $$i=1.6$$
degree of dissociation/ionisation=$$\dfrac{i-1}{n-1}$$
$$NaCl \rightarrow Na^+ + Cl^-$$ So, $$n=2$$
$$\alpha =0.6$$
=$$60\%$$
Option A is correct.
Calculate solubility of $$Pb{I}_{2}$$ ($${K}_{sp}=1/4\times {10}^{-8}$$) in water at $${25}^{o}$$, which is $$90$$% dissociated.
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0%
$$7.5 \times 10^{-4} \ M$$
0%
$$1.6 \times 10^{-3} \ M$$
0%
$$3.4 \times 10^{-3} \ M$$
0%
$$8.2 \times 10^{-3} \ M$$
Water and chlorobenzene are immiscible liquids. Their mixture boils at $$89^o$$C under a reduced pressure of $$7.7 \times 10^4$$ Pa. The vapour pressure of pure water at $$89^o$$C is $$7 \times 10^4$$ Pa. Weight percent of chlorobenzene in the distillate is:
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0%
50
0%
60
0%
79
0%
38.46
Explanation
Vapour pressure = $$7.7 \times 10^4$$ Pa
Total pressure = $$7 \times 10^4$$ Pa
Partial pressure = Vapour pressure = Mole fraction x Total pressure
Mole fraction of $$H_2O$$ = $$\dfrac{Vapour pressure}{Total pressure}$$ =
$$\dfrac{7.7 \times 10^4}{7 \times 10^4}$$ = 0.909
Mole fraction of $$CHCl_3$$ = 1 - 0.909 = 0.091
Molar mass of $$H_2O$$ = 18g/mol
Molar mass of $$CHCl_3$$ = 119.5g/mol
Weight percent of chloroform =
Mole fraction of $$CHCl_3$$ x
Molar mass of $$CHCl_3$$ / [
Molar mass of $$CHCl_3$$ +
Molar mass of $$H_2O$$ }
Weight percent of chloroform = 79%
Answer is 79%.
The $$K_{sp}$$ for $$Cr(OH)_3$$ is $$1.6 \times 10^{-30}$$. The solubility of this compound in water is:
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0%
$$4 \sqrt{1.6\times10^{-30}}$$
0%
$$4 \sqrt{1.6\times10^{-30} / 27}$$
0%
$$1.6\times10^{-9}$$
0%
$$2 \sqrt{1.6\times10^{-30}}$$
Explanation
Let S M be the solubility of $$\displaystyle Cr(OH)_3$$
$$\displaystyle [Cr^{3+}]=[Cr(OH)_3]=S$$
$$\displaystyle [OH^-]=3[Cr(OH)_3]=3S$$
The solubility product $$\displaystyle K_{sp}= [Cr^{3+}][OH^-]^3$$
$$\displaystyle 1.6 \times 10^{-30} = S \times (3S)^3$$
$$\displaystyle 1.6 \times 10^{-30}=27S^4$$
$$\displaystyle S=[\dfrac {1.6 \times 10^{-30}}{27}]^{1/4}$$
$$\displaystyle S=3.6 \times 10^{-8} \ M$$
Hence, solubility of $$\displaystyle Cr(OH)_3$$ in water is $$\displaystyle 3.6 \times 10^{-8} \ M$$
The vapour pressure of the solution of two liquids $$A({p}^{o}=80mm$$) and $$B({p}^{o}=120mm)$$ is found to be $$100mm$$ when $${x}_{A}=0.4$$. The result shows that:
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solution exhibits ideal behaviour
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solution shows positive deviations
0%
solution shows negative deviations
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solution will show positive deviations for lower concentration and negative deviations for higher concentrations
Explanation
$$ \begin{aligned} P_{A}^{0} &=80 \mathrm{~mm} \quad P_{B}^{\circ}=120 \mathrm{~mm} \quad x_{n}=0.4 \quad x_{B}=0.6 \\ P_{T} &=x_{A} P_{A}^{0}+x_{B} P_{0}^{0} \\ &=0.4 \times 80+0.6 \times 120 \\ &=32+72 \\ P_{T} &=104 \mathrm{~} \end{aligned} $$
Since the given total pressure is less than the calculated Total pressure therefore, it shows a negative deviation hence option C is correct.
$$A$$ and $$B$$ form ideal solutions, at $${50}^{o}C$$, $${P}_{A}^{o}$$ is half $${P}_{B}^{o}$$. A solution containing $$0.2$$ mole of $$A$$ and $$0.8$$ mole of $$B$$ has a normal boiling point of $${50}^{o}C$$. Find $$18\times {P}_{B}^{o}$$. ($${P}_{B}^{o}$$ is in atm)
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0%
0.24
0%
0.34
0%
0.53
0%
0.46
The ratio of the vapour pressure of a solution to the vapour pressure of the solvent is:
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equal to the mole fraction of the solvent
0%
equal to the mole fraction of the solute
0%
directly proportional to mole fraction of the solute
0%
None of the above
Vapour Pressure of a mixture of benzene and toluene is given by $$P= 179X_{B} + 92$$, Where $$X_{B}$$ is mole fraction of benzene.
If Vapours are removed and condensed in to liquid then what will be the ratio of mole fraction of benzene and toluene in first condensate :
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0%
2.8
0%
1.5
0%
3.5
0%
4.5
Explanation
Given $$P= 179 X_B+92$$
For pure $$C_6H_6, X_B=1$$
$$\therefore P^o_B= 179+92=271mm$$
For pure $$C_7H_8, X_B=0$$
$$\therefore P^o_T= 179 \times 0+92=92mm$$
Now, $$P_M=P^o_BX_B+P^o_TX_T$$
$$=\left(271 \times \cfrac {12}{12+8}\right)+\left(92\times \cfrac {8}{12+8}\right)$$
$$= 199.4mm$$
as, Moles of $$C_6H_6= \cfrac {936}{78}=12$$
Moles of $$C_7H_8= \cfrac {736}{92}=8$$
Now, mole fraction of $$C_6H_6$$ in vapour phase of initial mixture $$X^1_B=\cfrac {162.6}{199.4}$$
and that of toulene, $$X^1_T= \cfrac {36.8}{199.4}$$
$$\therefore \cfrac {X^1_B}{X^1_T}=\cfrac {162.6}{36.8}=4.418$$
What is the solubility of $$Al(OH_{3}), K_{sp} = 1 \times 10^{-33}$$, in a solution having $$pH = 4$$?
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0%
$$6 \times 10^{-3}$$M
0%
$$10^{-6}$$M
0%
$$1.5 \times 10^{-4}$$M
0%
$$2.47 \times 10^{-9}$$M
Explanation
The solubility equilibrium of $$Al(OH)_3$$ is
$$Al(OH)_3$$ $${\rightarrow}_{\leftarrow}$$ $$Al^{+3}$$ + $$3OH^-$$
Therefore $$x=1,\ y=2$$ and
$$K_{sp}$$=$$[Al^{+3}]\times [OH^-]$$ = $$27S^4$$
$$S$$ = $$(\dfrac{10^{-33}}{27})^{1/4}$$
Therefore $$S$$= $$2.47\times 10^{-9}M$$ in $$ mol\ dm^{-3}$$
At a given temperature, total vapour pressure in Torr of a mixture volatile components A and B is given by
$$P_{Total}=120-75X_B$$
Hence, vapour pressure of pure A and B respectively (in Torr) are ?
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120, 75
0%
120, 195
0%
120, 45
0%
75, 45
Explanation
$$\displaystyle P_{Total}=P_AX_A+P_BX_B$$
But $$\displaystyle X_A=1-X_B$$
$$\displaystyle ( \because X_A+X_B=1)$$
Hence, $$\displaystyle P_{Total}=P_A(1-X_B)+P_BX_B$$
$$\displaystyle P_{Total}=P_A-P_AX_B+P_BX_B$$
$$\displaystyle P_{Total}=P_A-(P_A-P_B)X_B$$
But, $$\displaystyle P_{Total}=120-75X_B $$
Hence, $$\displaystyle P_A=120 \ torr$$
$$\displaystyle P_A-P_B = 75$$
$$\displaystyle P_B=P_A-75$$
$$\displaystyle P_B=120-75$$
$$\displaystyle P_B=45 \ torr$$
Vapour Pressure of a mixture of benzene and toluene is given by $$P= 179X_{B} + 92$$, Where $$X_{B}$$ is mole fraction of benzene.
Vapour pressure of the solution obtained by mixing 936 g of benzene and 736 gm of toluene :
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0%
199.4 mm
0%
271 mm
0%
280 mm
0%
289 mm
Explanation
P=179$$X_B$$+92
If [Benzene] =936g
If[Toluene]=736g
$$X_B$$= mole fraction of benzene in liquid
$$X_B$$=$$ \dfrac{\dfrac{936}{78}}{\dfrac{936}{78}+\dfrac{736}{92}}$$=0.6
Now$$ P_B$$=179$$\times0.6$$ +92
$$P_B$$=199.4
At $$80^0$$C, the vapour pressure of pure liquid A is $$250$$mm of Hg and that of pure liquid B is $$1000$$ mm of Hg. If a solution of A and B boils at $$80^0$$C and $$1$$ atm pressure, the amount of A in the mixture is :$$(1atm = 760mm$$Hg)
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0%
$$50$$ mole percent
0%
$$52$$ mole percent
0%
$$32$$ mole percent
0%
$$48$$ mole percent
Explanation
1 atm = 760 mm Hg = $$P_T$$
$$P_T = P^0_A X_A \times P^0_BX_B$$
$$760 = 250X_A + 1000(1 - X_A)$$
$$ 240 = 750X_A$$
$$X_A = 0.32$$
Hence mole $$\%$$ of A = $$32\%$$
The van't Hoff factor of the benzoic acid solution in benzene is 0.In this solution, benzoic acid:
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0%
dissociates
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forms dimer
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remains unchanged
0%
forms tetramer
Explanation
As colligative property $$\propto $$ $$n\ ({\text{no. of moles}})$$
Therefore, $$i\ ({\text{van't Hoff factor}})=\dfrac {\text{Observed number of particles}}{\text{Theoretical number of particles}}$$
But $$i=\dfrac {1}{2}$$, therefore, Observed number of particles <
Theoretical number of particles
Thus, the benzoic acid forms a dimer.
The vapour pressure of the solution of two liquids $$A(P^{\circ} = 80\ mm)$$ and $$B(p^{\circ} = 120\ mm)$$ is found to be $$100\ mm$$ when $$x_{A} = 0.4$$. The result shows that :
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0%
Solution exhibits ideal behaviour
0%
Solution shows positive deviations
0%
Solution shows negative deviations
0%
Solution will show positive deviations for lower concentration and negative deviations for higher concentrations
Explanation
$$P_A^o = 80\ mm$$
$$P_B^o = 120\ mm$$
$$x_A = 0.4 \,\,\,\, x_B = 0.6$$
$$P = 80 \times 0.4 + 120 \times 0.6$$
$$= 104\ mm$$
observed vapour pressure < calculated vapour pressure
(solution shows negative deviation
At $$80^o$$C, the vapour pressure of pure liquid $$A$$ is $$250$$ mm of $$Hg$$ and that of pure liquid $$B$$ is $$1000$$ mm of $$Hg$$. If a solution of $$A$$ and $$B$$ boils at $$80^o$$ and $$1$$ atm pressure, the amount of $$A$$ in the mixture is?
($$1$$ atm $$=760$$ mm Hg).
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0%
$$50$$ mole percent
0%
$$52$$ mole percent
0%
$$32$$ mole percent
0%
$$48$$ mole percent
Explanation
1 atm = 760 mm Hg = $$P_T$$
$$P_T = P^0_A X_A \times P^0_BX_B$$
$$760 = 250X_A + 1000(1 - X_A)$$
$$ 240 = 750X_A$$
$$X_A = 0.32$$
Hence mole $$\%$$ of A $$= 32\%$$
At 80C, the vapour pressure of pure benzene is 753 mm Hg and of pure toluene 290 mm Hg. Calculate the composition of a liquid in mole percent which at $$80$$ is in equilibrium with vapour containing 30 mole percent of benzene.
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0%
24, 56
0%
26 ,54
0%
56 24
0%
34 ,76
Explanation
Mole % of benzene=30%of solution
Moles of benzene=30*80/100=24
Moles of Toluene=70*80/100=56
Composition of benzene and Toluene are 24 moles and 56 moles respectively.
Two moles of pure liquid 'A' ($$P^o_A$$ = 80mm of Hg) and 3 moles of pure liquid 'B' ($$P^o_B$$ = $$120$$mm of $$Hg$$) are mixed. Assuming ideal behaviour.
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Vapour pressure of the mixture is $$104$$ mm of $$Hg$$
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Mole fraction of liquid '$$A'$$ in Vapour pressure is $$0.3077$$
0%
Mole fraction of '$$B$$' in vapour pressure is $$0.692$$
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Mole fraction of '$$B$$' in vapour pressure is $$0.785$$
Explanation
Formula of Ammonia = $$NH_{3}$$
$$P_{A}^{\circ}=80 mm Hg$$ $$n_{A}=2$$ mole
$$P_{B}^{\circ}=120 mm Hg$$ $$n_{B}=3$$ mole
$$x_{A}=\frac{n_{A}}{n_{A}+n_{B}}=\frac{2}{2+3}=0.4$$
$$x_{B}=\frac{n_{B}}{n_{A}+n_{B}}=\frac{3}{2+3}=0.6$$
Vapour pressure $$P_{A}=P_{A}^{\circ}x_{A}+P_{B}^{\circ}x_{B}$$
$$P_{A}=80\times 0.4+120\times 0.6$$
$$=104 mmHg$$
Option A is correct
Match the column I with column II and mark the appropriate choice
Column I
Column II
(A) Ethyl alcohol + water
(i) $$p={p}^{o}x$$
(B) Benzene + Toulene
(ii) Effect of pressure on gas solutions
(C) Henry's law
(iii) Ideal solution
(D) Raoult's law
(iv) Azeotropic mixture
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(A) $$\rightarrow$$ (i); (B) $$\rightarrow$$ (ii); (C)$$\rightarrow$$ (iii); (D) $$\rightarrow$$ (iv)
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(A) $$\rightarrow$$ (i); (B) $$\rightarrow$$ (iii); (C)$$\rightarrow$$ (ii); (D) $$\rightarrow$$ (iv)
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(A) $$\rightarrow$$ (iv); (B) $$\rightarrow$$ (iii); (C)$$\rightarrow$$ (ii); (D) $$\rightarrow$$ (i)
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(A) $$\rightarrow$$ (iii); (B) $$\rightarrow$$ (i); (C)$$\rightarrow$$ (i); (D) $$\rightarrow$$ (iv)
Explanation
Column I
Column II
(A) Ethly alcohol $$+$$ water $$-$$ Azerotropic mixture
(B) Benzene $$+$$ Toluene $$-$$ Ideal solution
(C) Henry's law $$-$$ Effect of pressure on gas solutions
(D) Raolt's law $$-$$ $$p={ p }^{ 0 }x$$.
What are the conditions for an ideal solution which obeys Raoult's law over the entire range of concentration?
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$${ \Delta }_{ mix }H=0,{ \Delta }_{ mix }V=0,{ P }_{ Total }={ p }_{ A }^{ o }{ x }_{ A }+{ p }_{ B }^{ o }{ x }_{ B }$$
0%
$${ \Delta }_{ mix }H=+ve,{ \Delta }_{ mix }V=0,{ P }_{ Total }={ p }_{ A }^{ o }{ x }_{ A }+{ p }_{ B }^{ o }{ x }_{ B }$$
0%
$${ \Delta }_{ mix }H=0,{ \Delta }_{ mix }V=+ve,{ P }_{ Total }={ p }_{ A }^{ o }{ x }_{ A }+{ p }_{ B }^{ o }{ x }_{ B }$$
0%
$${ \Delta }_{ mix }H=0,{ \Delta }_{ mix }V=0,{ P }_{ Total }={ p }_{ B }^{ o }{ x }_{ B }$$
Explanation
For an ideal solution that obeys Raoult's Law the following are valid:
Heat is neither released nor absorbed during the reaction, $$\Delta_{mix}H=0$$
The volume of the solution remains the same $$\Delta_{mix}V=0$$
For an ideal solution, $$\Delta H$$ and$$\Delta V$$ for mixing should be zero.
$${ P }_{ Total }={ p }_{ A }+{ p }_{ B }$$ and A-A, B-B and A-B interactions should be nearly the same.
For which of the following solutes the van't Hoff factor is not greater than one?
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0%
$$Na{NO}_{3}$$
0%
$$Ba{Cl}_{2}$$
0%
$${ K }_{ 4 }\left[ Fe{ \left( CN \right) }_{ 6 } \right] $$
0%
$${NH}_{2}CO{NH}_{2}$$
Explanation
$${ NH }_{ 2 }{ CONH }_{ 2 }\rightarrow $$ urea
urea is a non-electrolyte and for non-electrolytes $$i=1$$
$$i=1-\alpha +n\alpha $$
$$i=1$$ {$$\alpha =0$$ for non-electrolytes}
A solute $$X$$ when dissolved in a solvent associate to form a pentamer. The value of van't Hoff factor ($$i$$) for the solute will be:
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0%
$$0.5$$
0%
$$5$$
0%
$$0.2$$
0%
$$0.1$$
Explanation
$$nA\longrightarrow { \left( A \right) }_{ n }$$
initially $$1 mol$$ $$0$$
after assoociation $$1-\alpha $$ $$\alpha /n$$
$$i=1-\alpha +\dfrac { \alpha }{ n } $$
$$i=1-1+\dfrac { 1 }{ n } $$ ($$\because$$ $$\alpha=1$$)
$$i=\dfrac { 1 }{ 5 } =0.2$$
What will be the correct order of vapour pressure of water, acetone and ether at $$30^0$$C. Given that among these compounds, water has maximum boiling point and ether has minimum boiling point.
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0%
Water < Ether < Acetone
0%
Water < Acetone < Ether
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Ether < Acetone < Water
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Acetone < Ether < Water
Explanation
Higher the boiling point lower will be the vapour pressure. Thus, the correct order of vapour pressure is Water < Acetone < Ether.
If $$\alpha$$ is the degree of dissociation of $${Na}_{2}{SO}_{4}$$, the vant Hoff's factor ($$i$$) used for calculating the molecular mass is:
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0%
$$1+\alpha$$
0%
$$1-\alpha$$
0%
$$1+2\alpha$$
0%
$$1-2\alpha$$
Explanation
For $${ Na }_{ 2 }{ SO }_{ 4 }:-i=n$$
$${ Na }_{ 2 }{ SO }_{ 4 }\longrightarrow \underbrace { { 2Na }^{ + }+{ SO }_{ 4 }^{ 2- } } $$
$$n=2$$
$$i=1-\alpha +n\alpha $$
$$i=1-\alpha +2\alpha $$
$$i=1+2\alpha $$
$$6 g$$. of the area of dissolved in $$90 g$$. of boiling water. The vapour pressure of the solution is:
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0%
$$744.8$$ mm
0%
$$758$$ mm
0%
$$761$$ mm
0%
$$760$$ mm
Explanation
Vapour pressure $$P=P°x,\quad P°=$$Boiling point of pure solvent
$$P=P°\cfrac { { n }_{ 1 } }{ { n }_{ 1 }+{ n }_{ 2 } }$$
$$=760\left( \cfrac { 0.1 }{ 0.1+5 } \right) \quad \quad 90g{ H }_{ 2 }O=5mol={ n }_{ 2 };\quad 6g$$ urea$$=0.1mol={ n }_{ 1 }$$
$$=760\left( \cfrac { 0.1 }{ 0.1+5 } \right) $$
$$=760\times 0.0196=14.89 $$ Torr=lowered pressure
Now, 1 torr = 1 mm of Hg
Vapour pressure of solution$$=760-14.89\approx 744.8 mm\ of\ Hg$$
Option A is correct.
Among the following substances, the lowest vapour pressure is exerted by:
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water
0%
alcohol
0%
ether
0%
mercury
Explanation
Vapor pressure of mercury is very low (compared to a liquid like methyl alcohol) because the forces of interaction between the individual metal atoms of mercury is quite a bit stronger than the cohesive molecular forces (such as a hydrogen bonding) that holds together several molecules in case of alcohols,ethers and water.
The van't Hoff factor of $$0.005M$$ aqueous solution of $$KCl$$ is $$1.95$$. The degree of ionisation of $$KCl$$ is:
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0%
$$0.95$$
0%
$$0.97$$
0%
$$0.94$$
0%
$$0.96$$
Explanation
The van't Hoff factor
is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its
mass. For most non-electrolytes dissolved in water, the van't Hoff factor
is essentially 1.
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Practice Class 12 Engineering Chemistry Quiz Questions and Answers
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