✒️FREEZING-POINT DEPRESSION
[tex]\tt{\huge{\blue{Solution:}}}[/tex]
The problem asks to solve for the molecular weight of the solute. First, we need to identify the quantities to be used. These are
m = molality
mass = mass of solute
n = number of moles of solute
MW = molecular weight of solute
[tex]\text{kg}_{solv} =[/tex] kilograms of solvent
[tex]K_{\text{f}} =[/tex] molal freezing-point depression constant
[tex]\Delta T_{\text{f}} =[/tex] freezing-point depression
[tex]T{{\circ}\atop{\text{f}}} =[/tex] freezing point of the pure solvent
[tex]T_{\text{f}} =[/tex] freezing point of the solution
Second, we need to identify the given values. These are
mass = 0.359 g
[tex]\text{kg}_{solv} =[/tex] 43.0 g = 0.0430 kg
[tex]K_{\text{f}} =[/tex] 5.12 °C/m
[tex]T{{\circ}\atop{\text{f}}} =[/tex] 5.5°C
[tex]T_{\text{f}} =[/tex] 5.15°C
Third, we need to find the freezing-point depression of the solution. The freezing-point depression is
[tex]\Delta T_{\text{f}} = T{{\circ}\atop{\text{f}}} - T_{\text{f}}[/tex]
[tex]\Delta T_{\text{f}} = 5.5^{\circ}\text{C} - 5.15^{\circ}\text{C}[/tex]
[tex]\Delta T_{\text{f}} = 0.35^{\circ}\text{C}[/tex]
Fourth, we need to find the molality of the solution. The molality is
[tex]m = \dfrac{\Delta T_{\text{f}}}{K_{\text{f}}}[/tex]
[tex]m = \dfrac{0.35^{\circ}\text{C}}{5.12^{\circ}\text{C/m}}[/tex]
m = 0.068359 m
Fifth, we need to find the number of moles of solute. The number of moles of solute is
[tex]n = m(\text{kg}_{solv})[/tex]
[tex]n = (0.068359 \: m)(\text{0.0430 kg})[/tex]
n = 0.0029395 mol
Finally, we can now solve for the molecular weight of solute. Therefore, the molecular weight of solute is
[tex]\text{MW} = \dfrac{\text{mass}}{n}[/tex]
[tex]\text{MW} = \dfrac{\text{0.359 g}}{\text{0.0029395 mol}}[/tex]
[tex]\boxed{\text{MW = 122 g/mol}}[/tex]
[tex]\\[/tex]
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