Question

In an experiment on depression of freezing point, two beakers leveled as A and B were filled with 1000 g. of water each. 40 g and 80 g of a certain non-electrolytic solute were added to both the beakers and freezing point of the solutions were noted as $T_{fA}$ and $T_{fB}$ respectively. Now both the solutions were poured into an empty beaker C and the freezing point of the mixed solution is found to be $T_{fC}$ . What is the ratio $T_{fA}:T_{fB}:T_{fC}?$

• A. 1 : 2 : 3
• B. 2 : 4 : 3
• C. 2 : 4 : 8
• D. 1 : 4 : 8

Answer 1

The correct option is B 2 : 4 : 3

$△T_f=K_{f}m$
$\Rightarrow 0-T_f=K_{f}m$
$\Rightarrow =-K_{f}m$
If 40 g of solute contains 'x'~moles of it then 80 g of the solut ill contains '2x' moles of it
$m_A=\frac{x}{\frac{Wt.ofsolventing}{1000}}=\frac{x}{\frac{1000}{1000}}=x$
$m_B=\frac{2x}{\frac{Wt.ofsolventing}{1000}}=\frac{x}{\frac{1000}{1000}}=2x$
$m_C=\frac{3x}{\frac{Wt.ofsolventing}{1000}}=\frac{3x}{\frac{1000}{1000}}=1.5x$
Required ration $T_{fA}:T_{fB}:T_{fc}=1.2:1.5=2:4:3$