Question

Molality of an aqueous solution of urea is 4.44 m. Mole fraction of urea in solution is x × 10–3. Value of x is _______. (integer answer)

Answer 1

Molality ($m$) of urea is given as 4.44 $m$, meaning 4.44 moles of urea are dissolved in 1000 g of water.
Step 1: Mole fraction formula
$X_{\text{urea}} = \frac{\text{Moles of urea}}{\text{Moles of urea} + \text{Moles of water}}$
Step 2: Calculate moles of water
$\text{Mass of water} = 1000 \, \text{g}, \quad \text{Molar mass of water} = 18 \, \text{g/mol}.$ $\text{Moles of water} = \frac{1000}{18} = 55.56.$
Step 3: Substitute values into the mole fraction formula
$X_{\text{urea}} = \frac{4.44}{4.44 + 55.56}.$ $X_{\text{urea}} = \frac{4.44}{60.00} = 0.0740.$
Step 4: Express mole fraction as $x \times 10^{-3}$
$X_{\text{urea}} = 74 \times 10^{-3}.$ $x = 74.$
Final Answer: 74

Answer 2

Molality ($m$) of urea is given as 4.44 $m$, meaning 4.44 moles of urea are dissolved in 1000 g of water.
Step 1: Mole fraction formula
$X_{\text{urea}} = \frac{\text{Moles of urea}}{\text{Moles of urea} + \text{Moles of water}}$
Step 2: Calculate moles of water
$\text{Mass of water} = 1000 \, \text{g}, \quad \text{Molar mass of water} = 18 \, \text{g/mol}.$ $\text{Moles of water} = \frac{1000}{18} = 55.56.$
Step 3: Substitute values into the mole fraction formula
$X_{\text{urea}} = \frac{4.44}{4.44 + 55.56}.$ $X_{\text{urea}} = \frac{4.44}{60.00} = 0.0740.$
Step 4: Express mole fraction as $x \times 10^{-3}$
$X_{\text{urea}} = 74 \times 10^{-3}.$ $x = 74.$
Final Answer: 74