Molality of an aqueous solution of urea is $4.44 \mathrm{~m}$. Mole fraction of urea in solution is $x \times 10^{-3}$, Value of $x$ is ________. (Integer answer)

<p>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.</p> <p>Molality ($\mathrm{m}$) is given by:</p> <p>$\mathrm{m} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$</p> <p>Here, the molality is $4.44 \ \mathrm{m}$, which means there are $4.44$ moles of urea (solute) per kilogram of water (solvent).</p> <p>The next step involves calculating the mole fraction. Mole fraction ($X$) of urea is given by:</p> <p>$$ X_{\text{urea}} = \frac{\text{moles of urea}}{\text{moles of urea} + \text{moles of water}} $$</p> <p>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:</p> <p>$$ \text{Moles of water} = \frac{1000 \ \mathrm{g}}{18 \ \mathrm{g/mol}} \approx 55.56 \ \text{moles} $$</p> <p>We then substitute these values into the mole fraction formula:</p> <p>$X_{\text{urea}} = \frac{4.44}{4.44 + 55.56}$</p> <p>Simplifying the expression inside the denominator first:</p> <p>$4.44 + 55.56 = 60$</p> <p>So, the mole fraction of urea becomes:</p> <p>$X_{\text{urea}} = \frac{4.44}{60}$</p> <p>Dividing the values gives us:</p> <p>$X_{\text{urea}} \approx 0.074$</p> <p>The problem asks for the mole fraction in the form $x \times 10^{-3}$, so we convert our result to this form:</p> <p>$X_{\text{urea}} = 74 \times 10^{-3}$</p> <p>Therefore, the value of $x$ is:</p> <p>$\boxed{74}$</p>