Question:
3.1 g of a compound, “X” (molar mass = 62 g mol\(^{-1}\)) is dissolved in 19.5 g of another compound, Y (molar mass = 78 g mol\(^{-1}\)). The ratio of mole fractions of X and Y in the solution is:
3.1 g of a compound, “X” (molar mass = 62 g mol\(^{-1}\)) is dissolved in 19.5 g of another compound, Y (molar mass = 78 g mol\(^{-1}\)). The ratio of mole fractions of X and Y in the solution is:
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Mole fraction ratio = moles of X / moles of Y. Always convert mass to moles using molar mass.
Updated On: Jun 26, 2026
- 1:5
- 5:1
- 4:1
- 1:4
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The Correct Option is A
Solution and Explanation
Step 1: Calculate moles of X.
\[ n_X = \frac{3.1}{62} = 0.05 \text{ mol} \]
Step 2: Calculate moles of Y.
\[ n_Y = \frac{19.5}{78} = 0.25 \text{ mol} \]
Step 3: Calculate mole fractions.
\[ x_X = \frac{n_X}{n_X + n_Y} = \frac{0.05}{0.05 + 0.25} = 0.05/0.3 = 1/6 \] \[ x_Y = \frac{n_Y}{n_X + n_Y} = 0.25/0.3 = 5/6 \]
Step 4: Ratio of mole fractions.
\[ x_X : x_Y = 1:5 \]
Step 5: Conclusion.
\[ \boxed{1:5} \]
\[ n_X = \frac{3.1}{62} = 0.05 \text{ mol} \]
Step 2: Calculate moles of Y.
\[ n_Y = \frac{19.5}{78} = 0.25 \text{ mol} \]
Step 3: Calculate mole fractions.
\[ x_X = \frac{n_X}{n_X + n_Y} = \frac{0.05}{0.05 + 0.25} = 0.05/0.3 = 1/6 \] \[ x_Y = \frac{n_Y}{n_X + n_Y} = 0.25/0.3 = 5/6 \]
Step 4: Ratio of mole fractions.
\[ x_X : x_Y = 1:5 \]
Step 5: Conclusion.
\[ \boxed{1:5} \]
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