The relative lowering of vapour pressure caused by dissolving 71.3 g of a substance in 1000 g of water is `7.13xx10^(-3)`. The molecular mass of the substance is (consider the solution is highly diluted )

Options

A. 180 g/mol
B. 18 g/mol
C. 1.8 g/mol
D. 360 g/mol

Answer

To find the molecular mass of the substance given the relative lowering of vapor pressure, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Relative Lowering of Vapor Pressure**:
The relative lowering of vapor pressure (ΔP/P₀) is given as \( 7.13 \times 10^{-3} \). This can also be expressed as:
\[ \text{Relative lowering of vapor pressure} = \frac{\Delta P}{P_0} = x_2 \]
where \( x_2 \) is the mole fraction of the solute.

2. **Finding the Mole Fraction of the Solvent**:
The mole fraction of the solvent (water) can be calculated as:
\[ x_1 = 1 - x_2 \]
Substituting the value of \( x_2 \):
\[ x_1 = 1 - 7.13 \times 10^{-3} = 0.99287 \]

3. **Expressing Mole Fraction**:
The mole fraction of the solvent can also be expressed in terms of moles:
\[ x_1 = \frac{\text{moles of solvent}}{\text{total moles}} = \frac{n_1}{n_1 + n_2} \]
where \( n_1 \) is the number of moles of solvent (water) and \( n_2 \) is the number of moles of solute (substance).

4. **Calculating Moles of Solvent**:
The number of moles of water (solvent) can be calculated using its mass and molar mass:
\[ n_1 = \frac{\text{mass of water}}{\text{molar mass of water}} = \frac{1000 \, \text{g}}{18 \, \text{g/mol}} \approx 55.56 \, \text{mol} \]

5. **Setting Up the Equation**:
Now we can set up the equation:
\[ 0.99287 = \frac{n_1}{n_1 + n_2} \]
Rearranging gives:
\[ n_1 + n_2 = \frac{n_1}{0.99287} \]
Therefore:
\[ n_2 = \frac{n_1}{0.99287} - n_1 = n_1 \left( \frac{1}{0.99287} - 1 \right) \]

6. **Calculating Moles of Solute**:
Substituting \( n_1 \):
\[ n_2 = 55.56 \left( \frac{1}{0.99287} - 1 \right) \approx 55.56 \times 0.00713 \approx 0.396 \, \text{mol} \]

7. **Finding Molar Mass of the Solute**:
The molar mass of the solute can be calculated using the mass of the solute and the number of moles:
\[ \text{Molar mass} = \frac{\text{mass of solute}}{n_2} = \frac{71.3 \, \text{g}}{0.396 \, \text{mol}} \approx 180.56 \, \text{g/mol} \]

8. **Final Answer**:
Rounding off, the molecular mass of the substance is approximately:
\[ \text{Molecular mass} \approx 180 \, \text{g/mol} \]