The density of \( \mathrm{SO}_{2} \) and \( \mathrm{CO}_{2} \) are \( 2.9 \mathrm{~g} / \mathrm{dm}^{3} \) and \( 1.98 \mathrm{~g} / \mathrm{dm}^{3} \) respectively. How much faster does \( \mathrm{CO}_{2} \) diffuse through a porous pot?

To determine how much faster \( \mathrm{CO}_{2} \) diffuses through a porous pot compared to \( \mathrm{SO}_{2} \), we need to consider the relationship between the density of a gas and its diffusion rate. The diffusion rate of a gas is inversely proportional to its density. This means that the gas with the lower density will diffuse faster. Given: - Density of \( \mathrm{SO}_{2} \): \( 2.9 \, \mathrm{g} / \mathrm{dm}^{3} \) - Density of \( \mathrm{CO}_{2} \): \( 1.98 \, \mathrm{g} / \mathrm{dm}^{3} \) To find out how much faster \( \mathrm{CO}_{2} \) diffuses compared to \( \mathrm{SO}_{2} \), we can calculate the ratio of their densities. The ratio of the densities of \( \mathrm{CO}_{2} \) to \( \mathrm{SO}_{2} \) is: \[ \frac{1.98 \, \mathrm{g} / \mathrm{dm}^{3}}{2.9 \, \mathrm{g} / \mathrm{dm}^{3}} \] Let's calculate this ratio. Calculate the value by following steps: - step0: Calculate: \(\frac{1.98}{2.9}\) - step1: Convert the expressions: \(\frac{\frac{99}{50}}{\frac{29}{10}}\) - step2: Multiply by the reciprocal: \(\frac{99}{50}\times \frac{10}{29}\) - step3: Reduce the numbers: \(\frac{99}{5}\times \frac{1}{29}\) - step4: Multiply the fractions: \(\frac{99}{5\times 29}\) - step5: Multiply: \(\frac{99}{145}\) The ratio of the densities of \( \mathrm{CO}_{2} \) to \( \mathrm{SO}_{2} \) is approximately \( 0.682759 \) or \( \frac{99}{145} \). This means that \( \mathrm{CO}_{2} \) diffuses approximately 68.28% faster than \( \mathrm{SO}_{2} \) through a porous pot.