The freezing point depression constant for water is $1.86 ^\circ ext{C m}^{-1}$. If $5.00 ext{ g}$ $Na_2SO_

Question

The freezing point depression constant for water is $1.86 ^\circ	ext{C m}^{-1}$. If $5.00 	ext{ g}$ $Na_2SO_4$ is dissolved in $45.0 	ext{ g}$ $H_2O$, the freezing point is changed by $-3.82 ^\circ	ext{C}$. The Van’t Hoff factor for $Na_2SO_4$ is:

Accepted Answer

The depression in freezing point ($\Delta T_f$) for an electrolyte is given by the equation: $\Delta T_f = i 	imes K_f 	imes m$.

1.  **Calculate Molality ($m$):**
    *   Molar mass of $Na_2SO_4 = 2(23) + 32 + 4(16) = 142 	ext{ g mol}^{-1}$.
    *   Moles of solute ($n_2$) = $\frac{5.00 	ext{ g}}{142 	ext{ g mol}^{-1}} \approx 0.0352 	ext{ mol}$.
    *   Mass of solvent ($w_1$) = $45.0 	ext{ g} = 0.045 	ext{ kg}$.
    *   $m = \frac{0.0352 	ext{ mol}}{0.045 	ext{ kg}} \approx 0.782 	ext{ m}$.

2.  **Calculate Van't Hoff Factor ($i$):**
    *   Given $\Delta T_f = 3.82 ^\circ	ext{C}$ (magnitude of change) and $K_f = 1.86 ^\circ	ext{C m}^{-1}$.
    *   Rearranging the formula: $i = \frac{\Delta T_f}{K_f 	imes m}$.
    *   $i = \frac{3.82}{1.86 	imes 0.782} = \frac{3.82}{1.4545} \approx 2.626$.

Rounding to two decimal places, $i \approx 2.63$.

Step-by-Step Solution

Practice All CHEMISTRY Questions