Question

Calculate the mole fraction of glucose in an aqueous solution that contains 45 g of glucose in 45 g of water. Find out the weight of NaOH required to be disolved in 90 g of water in order to get a solution of the same mole fraction.

• A. $\frac{1}{11},20g$
• B. $\frac{1}{11},10g$
• C. $\frac{1}{10},10g$
• D. $\frac{1}{10},20g$

Answer 1

The correct option is A $\frac{1}{11},20g$

$MoleFraction=\frac{Molesofsolute}{Molesofsolute+Molesofsolvent}$

$Molesofglucose\left(solute\right)=\frac{45}{180}=0.25$ (Since molecular mass of $C_6{H}_{12}{O}_6$ i.e glucose is $180g$)

$Molesofsolvent=\frac{45}{18}=2.5$ (Since molecular mass of $H_{2}O$ i.e water is $18g$)

$Molefraction=\frac{0.25}{0.25+2.5}=0.0909=\frac{1}{11}$

Now, for mole fraction of $NaOH$, let us take weight of $NaOH$ to be dissolved $=xg$

So, $MolesofNaOH=\frac{x}{40}$ [Since, molecular weight of $NaOH$ is $40g$]

And $Molesof{H}_{2}O=\frac{90}{18}$

$Molefraction=\frac{x/40}{\frac{x}{40}+\frac{90}{18}}$

$0.0909=\frac{x/40}{\frac{x}{40}+\frac{90}{18}}$

Therefore, $x=20g$