Question

Calculate the freezing point of a 0.08500 m aqueous solution of nano3. the molal freezing-point-depression constant of water is 1.86°c/m. remember to include the value of i.

Answer 1

Depression in freezing point (Δ$T_{f}$) = $K_{f}$×m×i,
where, $K_{f}$ = cryoscopic constant = $1.86^{0} C/m$,
m= molality of solution = 0.0085 m
i = van't Hoff factor = 2 (For $NaNO_{3}$)

Thus, (Δ$T_{f}$) = 1.86 X 0.0085 X 2 = $0.03162^{0}C$

Now, (Δ$T_{f}$) = $T^{0}$ - T
Here, T = freezing point of solution
$T^{0}$ = freezing point of solvent = $0^{0}C$
Thus, T = $T^{0}$ - (Δ$T_{f}$) = -$0.03162^{0}C$

Answer 2

Answer : The freezing point of a solution is, $0.32^oC$

Explanation :

First we have to calculate the Van't Hoff factor (i) for $NaNO_3$.

The dissociation of $NaNO_3$ will be,

$NaNO_3\rightarrow Na^++NO_3^-$

So, Van't Hoff factor = Number of solute particles = $Na^++NO_3^-$ = 1 + 1 = 2

Now we have to calculate the freezing point of a solution.

Formula used for lowering in freezing point :

$\Delta T_f=i\times k_f\times m$

or,

$T_f^o-T_f=i\times k_f\times m$

where,

$\Delta T_f$ = change in freezing point

$T_f$ = temperature of solution = ?

$T^o_f$ = temperature of pure water = $0^oC$

$k_f$ = freezing point constant  = $1.86^oC/m$

m = molality  = 0.08500 m

i = Van't Hoff factor = 2

Now put all the given values in this formula, we get the freezing point of a solution.

$0^oC-T_f=2\times (1.86^oC/m)\times 0.08500m$

$T_f=0.32^oC$

Therefore, the freezing point of a solution is, $0.32^oC$