Question

When $.15$ g of a non-volatile solute was dissolved in $90$ g of benzene, the boiling point of benzene raised from $353.23$K to $353.93$ K. Calculate the molar mass of the solute.
($K_b$ for benzene $=2.52$ kg $mo{l}^{-1}$).

Answer 1

Elevation in boiling point of a solvent upon addition of solute is given as:

$\Delta T_b=m.K_b$

m=molality of the solution=number of moles of solute in 1 kg solvent

$m=\frac{m_s}{M_s\times m_B}$

$m_B=$mass of benzene=90g=0.09kg

$m_s=$mass of solute=0.15g

let molar mass of solute be $M_s$

thus $m=\frac{0.15}{M_s\times 0.09}$

$m=\frac{15}{9M_s}$

$K_b$=ebullioscopic constant$=2.52Kkg/mol$

Initial boiling point of benzene, $T_i=353.23K$

Final boiling point of solution, $T_f=353.93K$

Elevation in boiling point, $\Delta T=T_f-T_i$

$\therefore \Delta T_b=353.93-353.23$

or $\Delta T_b=0.7K$

substituting the values we get:

$\Delta T_b=m.K_b$

$0.7=\frac{15}{9M_s}\times 2.52$

$M_s=6g/mol$

Molar mass of the solute is 6 g/mol.