JEE Main 2024 — Some Basic Concepts Of Chemistry Question with Solution

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 ($${\mathrm{H}}_2\mathrm{O}$$) is approximately $$18g/mol$$. Therefore, the moles of water in $$1\mathrm{kg}$$ (or $$1000\mathrm{g}$$) of water are:

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

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:

$$\boxed{74}$$