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A solution of an unknown non-dissociating solute was prepared by dissolving

0.300 g of the substance in 40.0 g of CCl4. The boiling point of the solution was

0.357 °C higher than that of the pure solvent. Calculate the molar mass of the

solute. Kb=5.02 °C/m​

Sagot :

Given:

mass of solute = 0.300 g

mass of solvent = 40.0 g = 0.0400 kg

[tex]K_{\text{b}} = \text{5.02°C/molal}[/tex]

[tex]\Delta T_{\text{b}} = \text{0.357°C}[/tex]

solute: unknown non-dissociating solute

solvent: CCl₄

Required:

molar mass of solute

Solution:

Step 1: Calculate the molality of solution.

[tex]\text{molality} = \frac{\Delta T_{\text{b}}}{K_{\text{b}}}[/tex]

[tex]\text{molality} = \frac{\text{0.357°C}}{\text{5.02°C/molal}}[/tex]

molality = 0.0711155 molal

Step 2: Calculate the number of moles of solute.

Note: 1 molal = 1 mol/kg

moles of solute = molality × mass of solvent

moles of solute = 0.0711155 mol/kg × 0.0400 kg

moles of solute = 0.00284462 mol

Step 3: Calculate the molar mass of solute.

[tex]\text{molar mass of solute} = \frac{\text{mass of solute}}{\text{moles of solute}}[/tex]

[tex]\text{molar mass of solute} = \frac{\text{0.300 g}}{\text{0.00284462 mol}}[/tex]

[tex]\boxed{\text{molar mass of solute = 105.5 g/mol}}[/tex]

[tex]\\[/tex]

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