Question

the mole fraction of glucose in 0.1 molal aqueous solution of glucose will be

Answer 1

nice answer hope and 5

Answer 2

Answer : The mole fraction of glucose will be, $1.71\times 10^{-3}$

Solution :

As we are given 0.1 molal aqueous solution of glucose. That means, 0.1 moles of glucose present in 1 kilogram of water.

The number of moles of glucose = 0.1 moles

Now we have to calculate the moles of water.

$\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{1000g}{18g/mole}=55.5moles$

The number of moles of water = 55.5 moles

Now wee have to calculate the mole fraction of glucose.

$X_{glucose}=\frac{n_{glucose}}{n_{glucose}+n_{water}}$

where,

$X_{glucose}$ = mole fraction of glucose = ?

$n_{glucose}$ = moles of glucose = 0.1

$n_{water}$ = moles of glucose = 55.5

Now put all the given values in the above expression, we get the mole fraction of glucose.

$X_{glucose}=\frac{0.1}{0.1+55.5}=1.71\times 10^{-3}$

Therefore, the mole fraction of glucose will be, $1.71\times 10^{-3}$