Question
On mixing $$3\,g$$ of non - volatile solute in $$200\,mL$$ of water, its boiling point $$\left( {{{100}^ \circ }} \right)$$ becomes $${100.52^ \circ }C.$$ If $${K_b}$$ for water is $$0.6\,K/m$$ then molecular $$wt.$$ of solute is :
A.
$$10.5\,g\,mo{l^{ - 1}}$$
B.
$$12.6\,g\,mo{l^{ - 1}}$$
C.
$$15.7\,g\,mo{l^{ - 1}}$$
D.
$$17.3\,g\,mo{l^{ - 1}}$$
Answer :
$$17.3\,g\,mo{l^{ - 1}}$$
Solution :
$$\eqalign{
& \Delta {T_b} = {K_b}\frac{w}{M} \times \frac{{1000}}{W} \cr
& 0.52 = 0.6 \times \frac{3}{m} \times \frac{{1000}}{{200}}\left( {W = 200 \times 1)} \right) \cr
& m = \frac{{1.8 \times 5}}{{0.52}} = 17.3\,g\,mo{l^{ - 1}} \cr} $$