pA and pB are the vapour pressure of pure liquid components, A and B, respectively of an ideal binary solution. If xA represents the mole fraction of component A, the total pressure of the solution will be
A
$${p_A} + {x_A}\left( {{p_B} - {p_A}} \right)$$
B
$${p_A} + {x_A}\left( {{p_A} - {p_B}} \right)$$
C
$${p_B} + {x_A}\left( {{p_B} - {p_A}} \right)$$
D
$${p_B} + {x_A}\left( {{p_A} - {p_B}} \right)$$
Explanation
p = pAxA + pBxB
= pAxA + pB(1 – xA)
(As for binary solution xA + xB = 1)
= pAxA + pB – pBxA
= pB + xA(pA – pB)
2
AIPMT 2011 Mains
MCQ (Single Correct Answer)
A 0.1 molal aqueous solution of a weak acid is 30% ionized. If Kf for water is 1.86oC/m, the freezing point of the solution will be
A
$$-$$0.18oC
B
$$-$$0.54oC
C
$$-$$0.36oC
D
$$-$$0.24oC
Explanation
We know that $$\Delta $$Tf
= i × Kf × m
Here i is van’t Hoff’s factor.
i for weak acid is 1 + $$\alpha $$.
Here $$\alpha $$ is degree of dissociation i.e., 30/100 = 0.3
$$ \therefore $$ Tf
= i × Kf
× m = 1.3 × 1.86 × 0.1 = 0.24
$$ \therefore $$ Freezing point = – 0.24
3
AIPMT 2011 Prelims
MCQ (Single Correct Answer)
The freezing point depression constant for water is $$-$$1.86o C m$$-$$1. If 5.00 g Na2SO4 is dissolved in 45.0 g H2O, the freezing point is changed by $$-$$3.82oC. Calculate the van't Hoff factor for Na2SO4.
The van't Hoff factor $$i$$ for a compound which undergoes dissociation in one solvent and association in other solvent is respectively
A
less than one and greater than one
B
less than one and less than one
C
greater than one and less than one
D
greater than one and greater than one.
Explanation
From the value of van’t Hoff factor i it is
possible to determine the degree of dissociation
or association. In case of dissociation, i is greater
than 1 and in case of association i is less than 1.
Questions Asked from Solutions
On those following papers in MCQ (Single Correct Answer)
Number in Brackets after Paper Indicates No. of Questions