Question
A solution containing $$12.5\,g$$ of non-electrolyte substance in $$185\,g$$ of water shows boiling point elevation of $$0.80\,K.$$ Calculate the molar mass of the substance. $$\left( {{K_b} = 0.52\,K\,kg\,mo{l^{ - 1}}} \right)$$
A.
$$53.06\,g\,mo{l^{ - 1}}$$
B.
$$25.3\,g\,mo{l^{ - 1}}$$
C.
$$16.08\,g\,mo{l^{ - 1}}$$
D.
$$43.92\,g\,mo{l^{ - 1}}$$
Answer :
$$43.92\,g\,mo{l^{ - 1}}$$
Solution :
$$\eqalign{
& {M_B} = \frac{{{K_b} \times {W_B}}}{{\Delta {T_b} \times {W_A}}} \cr
& \,\,\,\,\,\,\,\,\, = \frac{{0.52 \times 12.5}}{{0.80 \times 0.185}} \cr
& \,\,\,\,\,\,\,\, = 43.92\,g\,mo{l^{ - 1}}\,\,\left( {\because \,\,185\,g = 0.185\,kg} \right) \cr} $$