Question

Equimolal solutions of $NaCl$ and $BaC{l}_2$ are prepared in water. Freezing point of $NaCl$ solution is found to be $2^{\circ }C$. Then the freezing point of $BaC{l}_2$ is: (Assume $100$% dissociation of $NaCl$ and $BaC{l}_2)$.

• A. $-2^{\circ }C$
• B. $-3^{\circ }C$
• C. $3^{\circ }C$
• D. $2^{\circ }C$

Answer 1

Answer: (B)

$-3^{\circ }C$

$\Delta T_b=K_b\times m\times i$

For $NaCl,$ $i=1+1=2$

$\therefore -2=K_b\times m\times 2$

$⟹K_b\times m=-1$

For $BaC{l}_2,$ $i=1+2=3$

$\therefore \Delta T_b=K_b\times m\times 3=-1\times 3=-3^{o}C$