Question

A solution contains 3.5 g of a non volatile solute in 125g of H2O and it boils at 373.52 k (Tb). Calculate the molar mass of the solute. [ Kb = 0.52K Kg/mol and Tb°= 373k]

Answer 1

we have the formula :

  Tb - Tb° = kb * m

  where m is the molality

 

molality = mass * 1000/( molar mass * mass of water)

m = 3.5 *1000/(M * 125)

m = 28/M

  where M is the molar mass of the solute

putting the values in the equation :

373.52 - 373 = 0.52 * (28/M)

0.52 = 0.52 * 28/M

M= 28

Answer 2

Given :

Mass of solute (m) = 3.5 g

Mass of solvent = 125 g

$T_b$° = 373 K

$T_b$ = 373.52 K

To Find :

Molar mass of the solute

Solution :

Elevation of boiling point of a dilute solution of a non-volatile, non-electrolyte solid solute is directly proportional to molal concentration of the solution.

                Δ$T_b$ ∝ m

or,             Δ$T_b$ =  $K_b$ × m

So, Elevation of Boiling Point (Δ$T_b$) = $K_b$ × molality (m)

⇒       (373.52 - 373)                          = 0.52 × $\frac{Moles Of Solute}{Weight of Solvent}$

⇒         0.52                                       = 0.52 × $\frac{(3.5/Molar Mass) * 1000}{125}$

⇒          125                                        =  $\frac{3500}{Molar Mass}$

∴    Molar Mass                                 =   28 gm

Hence, the molar mass of the solute is found to be 28 gm.