QJEEM24Ch9402: JEE Main 2024 Chemistry - Class 11 JEEM24Ch9402 Solution: Molality of an aqueous solution of urea is . Mole fraction of…

To determine the mole fraction of urea in a solution where the molality is $4.44 \ \mathrm{m}$ , we first need to understand the definitions and relationships involved.

Molality ( $\mathrm{m}$ ) is given by:

$\mathrm{m} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$

Here, the molality is $4.44 \ \mathrm{m}$ , which means there are $4.44$ moles of urea (solute) per kilogram of water (solvent).

The next step involves calculating the mole fraction. Mole fraction ( $X$ ) of urea is given by:

$X_{\text{urea}} = \frac{\text{moles of urea}}{\text{moles of urea} + \text{moles of water}}$

Since we have $4.44$ moles of urea, let's determine the moles of water. The molar mass of water ( $\mathrm{H_2O}$ ) is approximately $18 \ \mathrm{g/mol}$ . Therefore, the moles of water in $1 \ \mathrm{kg}$ (or $1000 \ \mathrm{g}$ ) of water are:

$\text{Moles of water} = \frac{1000 \ \mathrm{g}}{18 \ \mathrm{g/mol}} \approx 55.56 \ \text{moles}$

We then substitute these values into the mole fraction formula:

$X_{\text{urea}} = \frac{4.44}{4.44 + 55.56}$

Simplifying the expression inside the denominator first:

$4.44 + 55.56 = 60$

So, the mole fraction of urea becomes:

$X_{\text{urea}} = \frac{4.44}{60}$

Dividing the values gives us:

$X_{\text{urea}} \approx 0.074$

The problem asks for the mole fraction in the form $x \times 10^{-3}$ , so we convert our result to this form:

$X_{\text{urea}} = 74 \times 10^{-3}$

Therefore, the value of $x$ is:

$\boxed{74}$