Question

The freezing point of benzene C6H6 is 5.50°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in benzene is DDT . How many grams of DDT, C14H9Cl5 (354.5 g/mol), must be dissolved in 209.0 grams of benzene to reduce the freezing point by 0.400°C ?

Answer 1

Answer: 0.028 grams

Explanation:

Depression in freezing point :

Formula used for lowering in freezing point is,

$\Delta T_f=k_f\times m$

or,

$\Delta T_f=k_f\times \frac{\text{ Mass of solute in g}\times 1000}{\text {Molar mass of solute}\times \text{ Mass of solvent in g}}$

where,

$\Delta T_f$ = change in freezing point

$k_f$ = freezing point constant  (for benzene} =$5.12^0Ckg/mol$

m = molality

Putting in the values we get:

$0.400^0C=5.12\times \frac{\text{ Mass of solute in g}\times 1000}{354.5\times 209.0}$

${\text{ Mass of solute in g}}=0.028g$

0.028 grams of DDT (solute) must be dissolved in 209.0 grams of benzene to reduce the freezing point by 0.400°C.