Question

A solution was prepared by dissolving certain amount of compound in 31.8 g of CCl4 has a

boiling point of 0.392 K higher than that of pure CCl4. If the molar mass of compound is 128gmol-1

calculate the mass of the solute dissolved. [ given Kb for CCl4 = 5.02 K kg mol-1].​

Answer 1

The mass of the solute dissolved is 0.1378 g.

Explanation:

It is given that ;

The mass of solvent, CCl₄ $(W_{1} ) = 31.8 g$

The molar mass of solute, $(M_{2} ) = 128 g/mol$

The boiling point elevation constant $(K_{b} ) = 5.02 K.kg/mol$

The elevation in boiling point Δ$T_{b} = 0.392 K$

We need to find the mass of solute $(W_{2} ) = ?$

Applying the formula of Δ$T_{b}$,

$T_{b} = \frac{K_{b} * 1000 * W_{2} }{M_{2}*W_{1} }$

Rearranging the above formula to find W₂ (g);

$W_{2} = \frac{T_{b} * M_{2}*W_{1}}{1000*K_{b} }$

Substituting given values in the above formula,

$W_{2} = \frac{0.392 * 128 * 31.8}{1000*5.02}$

$W_{2} = 0.1378 g$

Thus, 0.1378 g of solute is dissolved in 31.8 g of CCl₄ to raise the boiling point by 0.392 K.