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 based on the given data, we can follow these steps:

### Step 1: Understand the formula for relative lowering of vapor pressure
The relative lowering of vapor pressure (ΔP/P₀) can be expressed as:
\[ \frac{P_0 - P}{P_0} = \frac{n_{solute}}{n_{solute} + n_{solvent}} \]
For a dilute solution, this can be simplified to:
\[ \frac{P_0 - P}{P_0} \approx \frac{n_{solute}}{n_{solvent}} \]
Where:
- \(P_0\) = vapor pressure of pure solvent
- \(P\) = vapor pressure of the solution
- \(n_{solute}\) = number of moles of solute
- \(n_{solvent}\) = number of moles of solvent

### Step 2: Calculate the number of moles of solvent (water)
Given that the mass of water is 1000 g and the molar mass of water (H₂O) is 18 g/mol:
\[ n_{solvent} = \frac{mass_{water}}{molar\ mass_{water}} = \frac{1000\ g}{18\ g/mol} \approx 55.56\ mol \]

### Step 3: Use the relative lowering of vapor pressure to find moles of solute
We know that the relative lowering of vapor pressure is given as \(7.13 \times 10^{-3}\):
\[ \frac{n_{solute}}{n_{solvent}} = 7.13 \times 10^{-3} \]
Thus, we can express the number of moles of solute as:
\[ n_{solute} = 7.13 \times 10^{-3} \times n_{solvent} = 7.13 \times 10^{-3} \times 55.56 \approx 0.396\ mol \]

### Step 4: Calculate the molar mass of the solute
We know the mass of the solute is 71.3 g. The molar mass (M) can be calculated using the formula:
\[ M = \frac{mass_{solute}}{n_{solute}} = \frac{71.3\ g}{0.396\ mol} \approx 180.56\ g/mol \]

### Step 5: Round to the nearest whole number
The molar mass of the substance is approximately 180.56 g/mol, which can be rounded to 180 g/mol.

### Final Answer
The molecular mass of the substance is approximately **180 g/mol**.

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