The vapour pressure of acetone at 20°C is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20°C, its vapour pressure was 183 torr. The molar mass (g mol

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Raoult Law and Vapour Pressure

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Question

The vapour pressure of acetone at 20°C is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20°C, its vapour pressure was 183 torr. The molar mass (g mol^–1) of the substance is :

488

128

$(D) \frac{P_{A}^{o} - P_{S}}{P_{S}} = \frac{n_{B}}{n_{A}}$

$\frac{185 - 183}{183} = \frac{1 .2 \times 58}{M \times 100}$

M = 64

This problem involves calculating the molar mass of a non-volatile solute using the relative lowering of vapor pressure. According to Raoult's Law for non-volatile solutes, the relative lowering of vapor pressure is proportional to the mole fraction of the solute.

Step 1: Understand Raoult's Law for non-volatile solutes

The formula for relative lowering of vapor pressure is:

\frac{P_{A}^{\circ} - P_{A}}{P_{A}^{\circ}} = x_{B}

Where:

$P_{A}^{\circ}$ is the vapor pressure of pure solvent (acetone) = 185 torr
$P_{A}$ is the vapor pressure of solution = 183 torr
$x_{B}$ is the mole fraction of the solute

Step 2: Calculate the relative lowering of vapor pressure

\frac{P_{A}^{\circ} - P_{A}}{P_{A}^{\circ}} = \frac{185 - 183}{185} = \frac{2}{185}

Step 3: Express mole fraction in terms of molar masses

The mole fraction of solute $x_{B}$ can be written as:

x_{B} = \frac{n_{B}}{n_{A} + n_{B}}

Where:

$n_{B}$ = moles of solute = $\frac{1.2}{M}$ (where M is the molar mass of solute)
$n_{A}$ = moles of solvent (acetone) = $\frac{100}{58}$ (molar mass of acetone, CH₃COCH₃ = 58 g/mol)

Since the solute is non-volatile and present in small amount, we can approximate:

x_{B} \approx \frac{n_{B}}{n_{A}} = \frac{\frac{1.2}{M}}{\frac{100}{58}} = \frac{1.2 \times 58}{100 \times M}

Step 4: Equate and solve for M

\frac{2}{185} = \frac{1.2 \times 58}{100 \times M}

Now cross-multiply and solve for M:

2 \times 100 \times M = 185 \times 1.2 \times 58

200 M = 185 \times 69.6

200 M = 12876

M = \frac{12876}{200} = 64.38 \approx 64 {g mol}^{- 1}

Final Answer: The molar mass of the substance is 64 g mol^–1.

Important Formulae

Relative lowering of vapor pressure:

\frac{P^{\circ} - P}{P^{\circ}} = x_{2} \approx \frac{n_{2}}{n_{1}}

Where $P^{\circ}$ is vapor pressure of pure solvent, P is vapor pressure of solution, $x_{2}$ is mole fraction of solute, $n_{1}$ and $n_{2}$ are moles of solvent and solute respectively.

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Important Formulae

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