the boiling point of benzene is 353. 23k when 1.80 gram non volatile solute was dissolved is 90 gram of benzene the boiling point raise to 354.11 k calculate molar mass of solute kb for benzene is 2.53 kg per mole

Given,

Weight of solute = 1.80g

Weight of solvent = 90 g

K_b for benzene = 2.53 kg/ mole

Suppose Molar weight of solute = w g

Boiling point elevation = 354.11 K - 353.23 K = 0.88 K

T_b =K_b * M  
and
$M = \frac{1.80 x 1000}{w x 90}$

0.88 =$\frac{2.53 x 1.80 x 1000}{w x 90}$

w =$\frac{2.53 x 1.80 x 1000}{0.88 x 90}$

$w = \frac{4554}{79.2} = 57.5$
Hence the weight of the solute is 57.5 g
Regards