Calculate the van't Hoff factors of the following weak electrolyte solutions: (a) \(0.050 \mathrm{m}\) HCHO \(_{2}\), which begins to freeze at \(-0.0986^{\circ} \mathrm{C}\).(b) \(0.100 \mathrm{M} \mathrm{HNO}_{2},\) which has a hydrogen ion (and nitrite ion) concentration of \(6.91 \times 10^{-3} \mathrm{M}\).

Short Answer

Expert verified
The calculated van't Hoff factor for a 0.050m solution of HCHO_2 is computed and extracted from the result in Step 1. And for a 0.100M solution of HNO_2, the calculated van't Hoff factor is computed and extracted from the result in Step 2.

Step by step solution

01

Calculation of van't Hoff factor for part (a)

The freezing point depression of the solution is \(\Delta T_{f} = 0.0986^{\circ}C\) and the molality (m) of the HCHO_2 solution is 0.050m. The cryoscopic constant for water \(K_{f}\) is 1.86 K·kg/mol. Now we rearrange the formula \(\Delta T_{f} = i \cdot K_{f} \cdot m\) for 'i': \(i = \frac{\Delta T_{f}}{K_{f} \cdot m}\). Now substitute the given values into the formula to calculate the van't Hoff factor.
02

Calculation of van't Hoff factor for part (b)

For HNO_2, the molar concentration is 0.100M and the ion concentration is \(6.91 \times 10^{-3} M\). The ion concentration indicates the concentration of both H+ ions and NO_2- ions, which totals to \(2 \times 6.91 \times 10^{-3} = 1.382 \times 10^{-2} M\). Van’t Hoff factor 'i' is calculated by dividing the molar ion concentration by the molar concentration of the compound: \(i = \frac{1.382 \times 10^{-2} M}{0.100 M}\). Calculate this ratio to obtain the van't Hoff factor.

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