Question

When $10.0\mathrm{g}$ of mercuric nitrate, $\mathrm{Hg}{\left({\mathrm{NO}}_3\right)}_2$, is dissolved in $1.00\mathrm{kg}$ of water, the freezing point of the solution is $-{0.162}^{\circ }\mathrm{C}$. When $10.0\mathrm{g}$ of mercuric chloride $\left({\mathrm{HgCl}}_2\right)$ is dissolved in $1.00\mathrm{kg}$ of water, the solution freezes at $-{0.0685}^{\circ }\mathrm{C}$. Use these data to determine which is the stronger electrolyte, $\mathrm{Hg}{\left({\mathrm{NO}}_3\right)}_2$ or ${\mathrm{HgCl}}_2$.

Short answer

The van't Hoff factors for mercuric nitrate and mercuric chloride are approximately 2.90 and 1.01, respectively. Since the van't Hoff factor for mercuric nitrate is higher, it indicates that it dissociates into more ions upon dissolving in water. Therefore, mercuric nitrate (Hg(NO3)2) is the stronger electrolyte compared to mercuric chloride (HgCl2).

Step-by-step solution

Finding the molality of each solution

Molality (m) is calculated as the moles of solute per kilogram of solvent. First, convert the grams of each solute to moles, then divide by the mass of the solvent in kilograms. For mercuric nitrate, Hg(NO3)2: Molar mass = 200.59 + 2*(14.01 + 3*16.00) = 324.6 g/mol Moles of Hg(NO3)2 = 10.0 g / 324.6 g/mol ≈ 0.0308 mol For mercuric chloride, HgCl2: Molar mass = 200.59 + 2*35.45 = 271.49 g/mol Moles of HgCl2 = 10.0 g / 271.49 g/mol ≈ 0.0368 mol Next, calculate the molality for both solutions: Molality_Hg(NO3)2 = 0.0308 mol / 1 kg ≈ 0.0308 mol/kg Molality_HgCl2 = 0.0368 mol / 1 kg ≈ 0.0368 mol/kg

Finding the van't Hoff factor for each compound

We can use the formula for freezing point depression to find the van't Hoff factor for each compound. The formula is: ΔTf = Kf * m * i We're given the freezing point depression (ΔTf) for both compounds and found their molalities in step 1. We also need the cryoscopic constant (Kf) for water, which is 1.86 °C kg/mol. Now, solve for the van't Hoff factor (i): i_Hg(NO3)2 = ΔTf_Hg(NO3)2 / (Kf * m_Hg(NO3)2) = (-0.162 °C) / (1.86 °C kg/mol * 0.0308 mol/kg) ≈ 2.90 i_HgCl2 = ΔTf_HgCl2 / (Kf * m_HgCl2) = (-0.0685 °C) / (1.86 °C kg/mol * 0.0368 mol/kg) ≈ 1.01

Comparing the van't Hoff factors

Now that we've found the van't Hoff factors for mercuric nitrate and mercuric chloride, we can compare them to see which compound acts as a stronger electrolyte. Since the van't Hoff factor for mercuric nitrate (i ≈ 2.90) is higher than that of mercuric chloride (i ≈ 1.01), it indicates that mercuric nitrate dissociates into more ions upon dissolving in water. This makes mercuric nitrate (Hg(NO3)2) the stronger electrolyte compared to mercuric chloride (HgCl2).