Question

What is the freezing point of a solution containing $\text{8}\text{.1 g HBr}$ in 100 g water assuming the acid to be 90% ionised? ( ${{k}_{f}}$ for water $=1.86\,\text{K}\,\text{mo}{{\text{l}}^{-1}}$ )

• A. $0.85{{\,}^{o}}C$
• B. $-3.53{{\,}^{o}}C$
• C. $0{{\,}^{o}}C$
• D. $-0.35{{\,}^{o}}C$

Answer 1

Answer: (B)

$-3.53{{\,}^{o}}C$

Answer 2

$\Delta {{\Tau }_{f}}=i\times {{k}_{f}}\times m$ $\text{Ions}\,\text{at}\,\text{equilibrium}\overset{HBr}{\mathop{1-\alpha }}\,\xrightarrow{{}}\underset{\alpha }{\mathop{{{H}^{+}}}}\,+\underset{\alpha }{\mathop{B{{r}^{-}}}}\,$ $\therefore$ $\text{Total}\,\text{ions}=1-\alpha +\alpha +\alpha$ $=1+\alpha$ $\therefore$ $i=1+\alpha$ Given, ${{k}_{f}}=1.86\,\text{K}\,\text{mo}{{\text{l}}^{-1}}$ mass of $\text{ }\\!\\!~\\!\\!\text{ HBr =}\,\text{8}\text{.1 g}$ mass of $\text{ }\\!\\!~\\!\\!\text{ }{{\text{H}}_{\text{2}}}\text{O}\,\text{=100 g}$ $\text{( }\\!\\!\alpha\\!\\!\text{ )}\,\,\text{=}$ degree of ionization $=90$ $m$ (molality) $\text{= }\frac{\text{mass}\,\text{of}\,\text{solute/mol}\text{.wt}\text{.of}\,\text{solute}}{\text{mass of solventin kg}}$ $=\frac{8.1/81}{100/1000}$ $i=1+\alpha$ $=1+90/100$ $=1.9$ $\Delta {{T}_{f}}=i\times {{k}_{f}}\times m$ $=1.9\times 1.86\times \frac{8.1/81}{100/1000}$ $=3.534{{\,}^{o}}C$ $\Delta {{T}_{f}}=$ (depression in freezing point) = freezing point of water - freezing point of solution 3.534 = 0-freezing point of solution $\therefore$ Freezing point of solution $=-3.534{{\,}^{o}}C$