The relative lowering of vapour pressure produced by dissolving 71.5 g of a substance in 1000 g of water is 0.00713. The molecular mass of the substance will be:

Options

A. 18.0 g/mol
B. 342 g/mol
C. 60 g/mol
D. 180 g/mol

Answer

To find the molecular mass of the substance, we can follow these steps:

### Step 1: Understand the formula for relative lowering of vapor pressure
The relative lowering of vapor pressure is given by the formula:
\[ \frac{P_0 - P}{P_0} = \text{mole fraction of solute} = \frac{n}{N} \]
where:
- \( P_0 \) = vapor pressure of pure solvent
- \( P \) = vapor pressure of the solution
- \( n \) = number of moles of solute
- \( N \) = number of moles of solvent

### Step 2: Express the mole fraction of solute
For dilute solutions, we can approximate the mole fraction of solute as:
\[ \text{mole fraction of solute} \approx \frac{n}{N} \approx \frac{\text{weight of solute}/M}{\text{weight of solvent}/M_w} \]
where:
- \( M \) = molecular mass of the solute (unknown)
- \( M_w \) = molecular mass of the solvent (for water, \( M_w = 18 \, \text{g/mol} \))

### Step 3: Substitute the known values
We know:
- Weight of solute = 71.5 g
- Weight of solvent = 1000 g
- Relative lowering of vapor pressure = 0.00713

Substituting these values into the equation gives:
\[ 0.00713 = \frac{(71.5/M)}{(1000/18)} \]

### Step 4: Rearranging the equation
Rearranging the equation to solve for \( M \):
\[ 0.00713 = \frac{71.5 \times 18}{1000 \times M} \]
\[ M = \frac{71.5 \times 18}{0.00713 \times 1000} \]

### Step 5: Calculate the molecular mass
Now we can calculate \( M \):
\[ M = \frac{71.5 \times 18}{0.00713 \times 1000} \]
Calculating the numerator:
\[ 71.5 \times 18 = 1287 \]
Calculating the denominator:
\[ 0.00713 \times 1000 = 7.13 \]
Now substituting back:
\[ M = \frac{1287}{7.13} \approx 180.504 \, \text{g/mol} \]

### Step 6: Conclusion
Thus, the molecular mass of the substance is approximately 180 g/mol.