Question

3.1 g of a compound, 'X' (molar mass = 62 g mol$^{-1}$) is dissolved in 19.5 g of other compound, Y (molar mass = 78 g mol$^{-1}$). The ratio of mole fractions of X and Y in the solution is

• A. 1:5
• B. 5:1
• C. 4:1
• D. 1:4

Hint:

Calculate moles: $n = \text{mass} / \text{molar mass}$.
Mole fraction of component $i$: $\chi_i = n_i / n_{total}$.
The ratio of mole fractions $\chi_X : \chi_Y$ is the same as the ratio of moles $n_X : n_Y$.
Double-check arithmetic, especially with decimals.

Answer 1

The Correct Option is D

First, calculate the number of moles for compound X and compound Y. Number of moles ($n$) = $\frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$. For compound X: Mass of X, $m_X = 3.1 \text{ g}$. Molar mass of X, $M_X = 62 \text{ g mol}^{-1}$. Number of moles of X, $n_X = \frac{m_X}{M_X} = \frac{3.1 \text{ g}}{62 \text{ g mol}^{-1}}$. $n_X = \frac{3.1}{62} = \frac{31}{620} = \frac{1}{20} = 0.05 \text{ mol}$. For compound Y: Mass of Y, $m_Y = 19.5 \text{ g}$. Molar mass of Y, $M_Y = 78 \text{ g mol}^{-1}$. Number of moles of Y, $n_Y = \frac{m_Y}{M_Y} = \frac{19.5 \text{ g}}{78 \text{ g mol}^{-1}}$. $n_Y = \frac{19.5}{78}$. To simplify $\frac{19.5}{78}$: Multiply numerator and denominator by 10: $\frac{195}{780}$. $195 \div 780$: Notice $78 \times 2 = 156$, $78 \times 2.5 = 156 + 39 = 195$. Or, $19.5 \times 4 = 78$. So, $\frac{19.5}{78} = \frac{1}{4} = 0.25 \text{ mol}$. So, $n_Y = 0.25 \text{ mol}$. Total number of moles in the solution, $n_{total} = n_X + n_Y$. $n_{total} = 0.05 \text{ mol} + 0.25 \text{ mol} = 0.30 \text{ mol}$. Mole fraction of X ($\chi_X$) = $\frac{n_X}{n_{total}} = \frac{0.05}{0.30} = \frac{5}{30} = \frac{1}{6}$. Mole fraction of Y ($\chi_Y$) = $\frac{n_Y}{n_{total}} = \frac{0.25}{0.30} = \frac{25}{30} = \frac{5}{6}$. (Check: $\chi_X + \chi_Y = 1/6 + 5/6 = 6/6 = 1$. Correct.) The question asks for the ratio of mole fractions of X and Y, i.e., $\chi_X : \chi_Y$. $\chi_X : \chi_Y = \frac{1}{6} : \frac{5}{6}$. Multiplying by 6, the ratio is $1 : 5$. This matches option (a). However, the provided image shows option (d) 1:4 as correct. Let me check if my calculation of $n_Y$ could lead to that. If $\chi_X : \chi_Y = 1:4$, then $\chi_X = 1/5$ and $\chi_Y = 4/5$. This would mean $n_X : n_Y = 1:4$. We have $n_X = 0.05$. If $n_X : n_Y = 1:4$, then $n_Y = 4 \times n_X = 4 \times 0.05 = 0.20 \text{ mol}$. For $n_Y = 0.20 \text{ mol}$: $m_Y = n_Y \times M_Y = 0.20 \times 78 = 15.6 \text{ g}$. But the given mass of Y is $19.5 \text{ g}$. So, my calculation of $n_Y = 0.25$ mol from $19.5 \text{g} / 78 \text{g/mol}$ is correct. The ratio $n_X : n_Y = 0.05 : 0.25$. To simplify this ratio, divide both by 0.05: $\frac{0.05}{0.05} : \frac{0.25}{0.05} = 1 : 5$. So, $\chi_X : \chi_Y = n_X : n_Y = 1:5$. My result 1:5 matches option (a). The marked answer (d) 1:4 is inconsistent with the provided data. I will proceed with my calculated answer. \[ \boxed{1:5} \] (Note: Solution leads to 1:5. Marked answer is 1:4, suggesting a possible typo in data or options/key.)