Question

What is the melting point of a solution in which 2.5 grams of sodium chloride is added to 230 mL of water?

Answer 1

0.976 should be the answer. 

Answer 2

$-0.347^{\circ} \mathrm{C}$ is the melting point of a solution in which 2.5 grams of sodium chloride is added to 230 mL of water

Explanation:

$\text {Density}=\frac{\text { mass }}{\text { volume }}$

$\text { mass of } H_{2} O=\text { density } \times \text {volume}$

$=\frac{1.0 g}{m L} \times 230 m L=230 g H_{2} O=0.230 \mathrm{kg} \mathrm{H}_{2} \mathrm{O}$

$\text {Moles solute } \mathrm{NaCl}=\frac{\text { mass }}{\text { molarmass }}=\frac{2.5 \mathrm{g}}{58.5 \mathrm{g} \text { per mol }}=0.0427 \mathrm{mol} \mathrm{NaCl}$

$K_{f}\text { of } H_{2} O \text { is } 1.86 \frac{^{\circ} \mathrm{C}}{m} \text { or } \mathrm{mol}^{-1} \mathrm{Kg}^{\circ} \mathrm{C}$

Melting is Solid changing into a liquid and it is just the reverse process of freezing

$\text { molality }(m)=\frac{\text { moles of solute }}{\text { mass of solvent in } k g}=\frac{0.0427 m o l}{0.230 k g}=0.187 m \text { or } 0.187 m o l K g^{-1}$

$\Delta T_{f}=-m \times K_{f}$

$T_{2}-0^{\circ}$ = -0.187 × 1.86

$T_{2}-0^{\circ}$ = $-0.347^{\circ} \mathrm{C}$

= $-0.347^{\circ} \mathrm{C}$ (Answer)