Question

A solution of 2.5 g of a non volatile solid in 100g benzene boiled at 0.42°c higher than the boiling point of pure benzene.Calculate the molecular mass of the substance.Molal elevation of 0.7k.Calculate the molecular mass of the substance.Molal elevation constant of benzene is 2.67k kg /mol.

Answer 1

may it would be helpful to you

Answer 2

The molecular mass of the substance is 159 g/mol.

Explanation:

Formula used for Elevation in boiling point :

$\Delta T_b=i\times k_b\times m$

or,

$T_b-T^o_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_b-T^o_b$ = $0.42^0C$

$k_b$ = boiling point constant  = $2.67K/m$

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte)

$w_2$ = mass of solute = 2.5 g

$w_1$ = mass of solvent (benzene) = 100 g

$M_2$ = molar mass of solute = ?

Now put all the given values in the above formula, we get:

$0.42^oC=1\times (2.67^oC/m)\times \frac{2.5g\times 1000}{M_2\times 100g}$

$M_2=159g/mol$

Learn more about elevation in boiling point

https://brainly.com/question/14325551

https://brainly.com/question/10714152