Question

Calculate the molality and mole fraction of the solute in aquous solution containing 3.0g of urea per 250g of water ?

Answer 1

Here is the answer of your question

Answer 2

The molality is 0.2 M and the mole fraction is 0.0035

Given:

Mass of urea or the solute = 3 g

Mass of the solvent = 250 g

We know that the mole mass of urea = 60 g

Mole mass of water = 18 g

Solution:

We can therefore calculate the molality of urea by using the formula given below:

$Molality =\frac{\text { Mass of solute } \times 1000}{\text { (Molar mass of solute } \times \text { Mass of solvent) }}$

$=\frac{3 \times 1000}{60 \times 250}$

$\Rightarrow \text {Molality}=0.2 \ M$

Now to calculate the mole fraction of urea in water, we use the formula given below:

$\text{Mole \ fraction \ of \ solute} = \frac{\text{Mole \ of \ solute}}{\text{Mole \ of \ solution}}$

Where, the mole of solution is given by adding together the moles of solute and solvent.

We know that the mass of urea is 3 g and the molar mass is 60 g which means that 60 g of urea contains 1 mole of urea.

So if 60 g urea = 1 mole of urea

$\text{3 g \ of \ urea} = \frac{3}{60} \ \text{moles}$

Similarly, 18 g of water = 1 mole of water

$\text{250 g of water} = \frac{250}{18} \ \text{moles of water}$

So, the moles of solution $=\frac{3}{60}+\frac{250}{18} = 13.93$

Mole fraction of urea $=\frac{3 / 60}{13.93}$

$=\frac{0.05}{13.93}$

$\Rightarrow \text { Mole fraction of urea }=0.0035$