Question

The boiling point of water $\left(100^{\circ} C \right)$ becomes $100.52^{\circ} C$, if $3$ grams of a non-volatile solute is dissolved in $200 \,ml$ of water. The moleular weight of solute is (if $K _{ b }$ for water is $=0.6\, K / m$ )

• A. $17.3\, g.mol ^{-1}$ :
• B. $15.4\, g.mol ^{-1}$ :
• C. $12.2\, g.mol ^{-1}$ :
• D. $20.4\, g.mol ^{-1}$ :

Answer 1

Answer: (A)

As deration in boilding point is given by:

$\Delta T=K_{b} \times$ molality

Putting the various values, we get:

$100.52-100.00=0.6 \times \frac{3 / M}{200 / 1000}$

$\Rightarrow M=17.3\, g\, mol ^{-1}$