Question

The depression in freezing point of 1 m NaCl solution, assuming NaCl to be 100% dissociated in water, is:

$\left(K_f=1.86K{m}^{-1}\right)$

• A. -1.86${}^{\circ }\text{C}$
• B. -3.72${}^{\circ }\text{C}$
• C. +1.86${}^{\circ }\text{C}$
• D. +3.72${}^{\circ }\text{C}$

Answer 1

The correct option is B -3.72${}^{\circ }\text{C}$

Since, NaCl is 100% dissociated,

$NaCl\left(aq\right)⟶N{a}^{+}\left(aq\right)+C{l}^{-}\left(aq\right)$

$\therefore$ $i=2$ for NaCl

we have,

$\Delta T_f=i{K}_{f}m$

$\Delta T_f=2\times 1.86\times 1$

so, Depression in freezing point of NaCl, $\Delta T_f=$ $-{3.72}^{0}C$