Question: 0.30 g of a volatile liquid displaces 90.0 $cm^3$ of air at STP in the Victor Meyer's method. The mo...

Question

0.30 g of a volatile liquid displaces 90.0 $cm^3$ of air at STP in the Victor Meyer's method. The molecular mass of the liquid is:

Answer 1

Solution

The Victor Meyer's method is used to determine the molecular mass of a volatile substance by measuring the volume of its vapor. In this method, the volume of air displaced by the vapor of the substance at a given temperature and pressure is measured. This displaced volume is equal to the volume occupied by the vapor of the substance under the same conditions.

Given: Mass of the volatile liquid (m) = 0.30 g Volume of air displaced at STP (V) = 90.0 cm³

At STP (Standard Temperature and Pressure), the molar volume of any ideal gas is 22.4 L/mol or 22400 cm³/mol. This means that 1 mole of any gas occupies 22400 cm³ at STP.

The volume of air displaced is 90.0 cm³ at STP, which means the volume of the vapor of the liquid is also 90.0 cm³ at STP.

We can find the number of moles (n) of the vapor using the molar volume at STP: $n = \frac{\text{Volume of vapor}}{\text{Molar volume at STP}}$ $n = \frac{90.0 \text{ cm}^3}{22400 \text{ cm}^3/\text{mol}}$

Now, we can calculate the molecular mass (M) of the liquid using the formula: $M = \frac{\text{Mass of liquid (m)}}{\text{Number of moles (n)}}$ Substitute the values of m and n: $M = \frac{0.30 \text{ g}}{\frac{90.0}{22400} \text{ mol}}$ $M = \frac{0.30 \text{ g} \times 22400 \text{ cm}^3/\text{mol}}{90.0 \text{ cm}^3}$ $M = \frac{6720}{90.0} \text{ g/mol}$ $M = 74.666... \text{ g/mol}$

Rounding the result to an appropriate number of significant figures (considering 0.30 g has two significant figures and 90.0 cm³ has three), we can round it to one decimal place, which gives 74.7 g/mol.

Comparing this value with the given options, the calculated value of 74.666... g/mol is closest to 74.8 g.