Question

$2.82g$ of glucose (molar mass$=180$) are dissolved in $30g$ of water. Calculate (a) the molarity (b) mole fraction of glucose and water.

Answer 1

We know that,

The weight of glucose$=2.82g$

Also,
no. of moles of glucose$=n_2=\frac{2.82}{180}=0.01567$
no. of moles of $N_{2}O=n_1=\frac{30}{18}=1.667$
Molality$=\frac{n_2}{n_1}\times \frac{1000}{m.wtofsolvent}$
$=\frac{0.01567}{1.667}\times \frac{1000}{18}=0.5222molal$
We have moles of glucose $\left(x_2\right)=0.01567$
moles of $H_{2}O\left(x_1\right)=1.667$
So, mole fraction of glucose
$x_2=\frac{n_2}{n_1+n_2}=\frac{0.0157}{\left(1.667+0.01567\right)}=0.0093$
mole fraction of $H_{2}O=x_1$
$\begin{array}{c}{x}_1=1-x_2=1=0.0093\\ x_1=0.9907\end{array}$