Question

30 ml of $C{H}_{3}OH(d=0.7980gc{m}^{-3})$ and $70ml$ of $H_{2}O(d=0.9984gc{m}^{-3})$ are mixed at ${25}^{\circ }C$ to form a solution of density $0.9575gc{m}^{-3}.$ $K_f\left(H_{2}O\right)$ is $1.86kgmo{l}^{-1}K$. Calculate its molarity (nearest integer).

Answer 1

Volume of $C{H}_{3}OH=V_1=30ml,d_1=0.798g/ml$

Thus, mass = density $\times$ volume = $m_1=30\times 0.798=23.94g$

Volume of $H_{2}O=V_2=70ml,d_2=0.9984g/ml$

Thus, mass = density $\times$ volume = $m_2=70\times 0.9984=69.888g$

Total mass of the solution = 23.94 + 69.888 g = $m_T=93.828g$

$d_{solution}=0.9575g/ml$

$V_{solution}=98ml$

$\Delta T_f=K_f\times molality=\frac{1.86\times 23.94\times 1000}{32\times 69.888}=19.91$

$T_f=-{19.91}^{\circ }C$

$Molarity=\frac{23.94}{32\times 0.098}=7.63M$

Hence answer is $8$.