(A) The number of moles of $X = \frac{W_X}{M_X} = \frac{3.1 \ g}{62 \ g \ mol^{-1}} = 0.05 \ mol$. The number of moles of $Y = \frac{W_Y}{M_Y} = \frac

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(A) The number of moles of $X = \frac{W_X}{M_X} = \frac{3.1 \ g}{62 \ g \ mol^{-1}} = 0.05 \ mol$. The number of moles of $Y = \frac{W_Y}{M_Y} = \frac{19.5 \ g}{78 \ g \ mol^{-1}} = 0.25 \ mol$. The ratio of mole fractions of $X$ and $Y$ is equal to the ratio of their number of moles: $\frac{\chi_X}{\chi_Y} = \frac{n_X}{n_Y} = \frac{0.05}{0.25} = \frac{1}{5} = 1:5$.

Class 12 Chemistry Solutions Method of expressing concentration of solution

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$3.1 \ g$ 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

AP EAMCET 2022 All AP EAMCET

A
$1:5$
B
$5:1$
C
$4:1$
D
$1:4$

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$3.1 \ g$ 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

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In a solution of a solute in benzene with a molality of $x \ m$,the mole fraction of the solute is $0.2$. The value of $x$ is ...... .

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