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

Step-by-step explanation:

Depression in freezing point :

Formula used for lowering in freezing point is,

`\Delta T_f=k_f* m`

or,

`\Delta T_f=k_f* \frac{\text{ Mass of solute in g}* 1000}{\text {Molar mass of solute}* \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* \frac{\text{ Mass of solute in g}* 1000}{354.5* 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.