Question

calculate the boiling point of a solution containing 0.456g of camphor mol.mass 152,dissolved in 31.4g of acetone boiling point=56.30ÂC,if the molecular elevation constant per 100g of acetone is 17.2Â.

Answer 1

Answer:The boiling point of a solution 56.46 °C.

Explanation:

Boiling point of acetone = T= 56.30 °C = 329.30 K

Boiling point of the solution with 0.456 g of camphor = $T_b$

$\Delta T_b=T_b-T$

$\Delta T_b=K_f\times m$

$K_f$ per 100 g of acetone = 17.2 K g/mol

$K_f$ 1000 g of acetone =1.72 K kg/mol

$\Delta T_b=1.72 K kg/mol\times \frac{0.456 g}{152 g/mol\times 0.0314 kg}=0.164 K$

$T_b=\Delta T_b+T=0.164 K+329.30 K=329.464 K$

$T_b=56.46^oC$

The boiling point of a solution 56.46 °C.

Answer 2

Answer:

The boiling point of a solution 56.46 °C.