$$ \therefore $$ Tf
= i × Kf
× m = 1.3 × 1.86 × 0.1 = 0.24
$$ \therefore $$ Freezing point = – 0.24
2
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.
4
AIPMT 2011 Mains
MCQ (Single Correct Answer)
200 mL of an aqueous solution of a protein contains its 1.26 g. The osmotic pressure of this solution at 300 K is found to be 2.57 $$ \times $$ 10$$-$$3 bar. The molar mass of protein will be (R = 0.083 L bar mol$$-$$1 K$$-$$1)