(D) Moles of water $= \frac{x}{18}$.Moles of ethanol $= \frac{69}{46} = 1.5$.Mole fraction of ethanol $(X_{ethanol}) = \frac{n_{ethanol}}{n_{water} +

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(D) Moles of water $= \frac{x}{18}$.Moles of ethanol $= \frac{69}{46} = 1.5$.Mole fraction of ethanol $(X_{ethanol}) = \frac{n_{ethanol}}{n_{water} + n_{ethanol}}$.Given $X_{ethanol} = 0.6$,so $0.6 = \frac{1.5}{\frac{x}{18} + 1.5}$.$\frac{x}{18} + 1.5 = \frac{1.5}{0.6} = 2.5$.$\frac{x}{18} = 2.5 - 1.5 = 1.0$.$x = 18 \, g$.

Class 12 Chemistry Solutions Method of expressing concentration of solution

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If the mole fraction of ethanol in a solution formed by mixing $x \, g$ of water and $69 \, g$ of ethanol is $0.6$,then $x =$ ...... $g$.

A
$54$
B
$36$
C
$180$
D
$18$

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If the mole fraction of ethanol in a solution formed by mixing $x \, g$ of water and $69 \, g$ of ethanol is $0.6$,then $x =$ ...... $g$.

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The molality of $648 \ g$ of pure water is ........... $m$.

$A$ solution of $Al_2(SO_4)_3$ $(d = 1.253 \ g/mL)$ contains $22\%$ salt by weight. The molarity,normality,and molality of the solution are:

Density of a $2.05 \ M$ solution of acetic acid in water is $1.02 \ g/mL$. The molality of the solution is ............. $mol \ kg^{-1}$.

Calculate the mass of urea $[NH_2CONH_2]$ required in making $2.5 \ kg$ of $0.25 \ m$ aqueous solution. (in $g$)

The molality and molarity of a solution of glucose in water which is labelled as $10 \% (w/w)$ are respectively (density of solution $= 1.2 \ g \ mL^{-1}$)

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