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 Xb0C Higher Than That Of The Pure Solvent. Calculate The Molar Mass Of The , Solute. Kb=5.02 Xb0C/m

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

Given:

mass of solute = 0.300 g

mass of solvent = 40.0 g = 0.0400 kg

$K_{\text{b}} = \text{5.02°C/molal}$

$\Delta T_{\text{b}} = \text{0.357°C}$

solute: unknown non-dissociating solute

solvent: CCl₄

Required:

molar mass of solute

Solution:

Step 1: Calculate the molality of solution.

$\text{molality} = \frac{\Delta T_{\text{b}}}{K_{\text{b}}}$

$\text{molality} = \frac{\text{0.357°C}}{\text{5.02°C/molal}}$

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.

$\text{molar mass of solute} = \frac{\text{mass of solute}}{\text{moles of solute}}$

$\text{molar mass of solute} = \frac{\text{0.300 g}}{\text{0.00284462 mol}}$

$\boxed{\text{molar mass of solute = 105.5 g/mol}}$

$\\$

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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 \xb0C higher than that of the pure solvent. Calculate the molar mass of the , solute. Kb=5.02 \xb0C/m

Given:

Required:

Solution:

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Given:

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Solution:

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