Question

The BP of benzene is 350.23k when 1.80 g of a non volulite solute was dissolved in 90 g of benzene the BP is raised to 354.11k. Calculate the molar mass of the solute kb for benzene is 2.53 k kg/mol.​

Answer 1

Given :

Boiling point of benzene = 350.23k

No of non volatile solute = 1.80 g

No. of benzene = 90 g

Boiling point of solution = 2.53 K kg/mol

To find :

The molar mass of the solute.

Solution :

1st we have to find elevation of B.P

The difference between boiling point of the solution and boiling point of the pure solvent is called elevation of boiling point.

$\boxed{\bf\Delta T_b = T_b - T_b ^\circ}$

by substituting the values in the formula,

$\dashrightarrow\sf\Delta T_b = T_b - T_b ^\circ$

$\dashrightarrow\sf\Delta T_b = 354.11 - 353.23$

$\dashrightarrow\sf\Delta T_b = 0.88$

we know that,

The relationship between molar mass of solute M₂ and elevation of boiling point is

$\boxed{ \bf M_2 = \dfrac{k_b \times W_2 \times 1000}{ \Delta T_b \times W_1}}$

by substituting the values in the formula,

$\dashrightarrow \sf M_2 = \dfrac{k_b \times W_2 \times 1000}{ \Delta T_b \times W_1}$

$\dashrightarrow \sf M_2 = \dfrac{2.53 \times 1.80\times 1000}{0.88 \times 90}$

$\dashrightarrow \boxed{\sf M_2 = 57.5 \: g \: mol ^{ - 1}}$

Thus, the molar mass of the solute is 57.5 g/mol.