# What is the freezing point of a 0.05500 m aqueous solution of NaNO_3, if the molal freezing-point-depression constant of water is 1.86°C /m?

##### 1 Answer
Dec 28, 2016

We know the depression of freezing point

$\Delta {T}_{f} = {K}_{f} \times b \times i$

Where

${K}_{f} \to \text{cryoscopic consrant"=1.86^@"C/m}$

$b \to \text{molal concentration} = 0.055 m$

$i \to \text{Van'tHoff factor"=2" for "NaNO_3 "*}$

$\text{*}$ As $N a N {O}_{3}$ dissociates as follows producing 2 ions per molecule.

$N a N {O}_{3} r i g h t \le f t h a r p \infty n s N {a}^{+} + N {O}_{3}^{-}$

So

$\Delta {T}_{f} = 1.86 \times 0.055 \times 2 \approx 0.2$

So the freezing point of the given aqueous solution will be $- {0.2}^{\circ} C$