If $5.85 g$ of $N a C l$ are dissolved in $90 g$ of water, the mole fraction of solute is ____________.

0.0196

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0.01

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0.1

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0.2

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Solution

The correct option is A $0.0196$
Given:-
Mass of $N a C l = 5.85 g$

Mass of $H_{2} O = 90 g$

Number of moles of $N a C l = \frac{5.85}{58.5} = 0.1$

Number of moles of $H_{2} O = \frac{90}{18} = 5$

Mole fraction of solute= $\frac{n_{solute}}{n_{solute} + n_{solvent}}$

Mole fraction= $\frac{0.1}{5 + 0.1}$

$\Rightarrow X_{B} = 0.0196$

Hence, the correct option is $A$

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