Question

Molality and mole-fraction of 3.0 g of urea (mol mass = 60) per 250 g water will be:-

• A. 0-2 m and 00-359
• B. 0-2 m and 0-00359
• C. 2.0 m and 0-0359
• D. 2.0 m and 0-00359.

Answer 1

Answer: (B)

0-2 m and 0-00359

Answer 2

Molality and mole fraction are concentration terms used in chemistry.

1. Calculate moles of solute (urea) and solvent (water):

• Moles of urea (solute):

  Given mass of urea = 3.0 g Molar mass of urea = 60 g/mol Moles of urea ($n_{\text{urea}}$) = $\frac{\text{Mass}}{\text{Molar Mass}} = \frac{3.0 \text{ g}}{60 \text{ g/mol}} = 0.05 \text{ mol}$

• Moles of water (solvent):

  Given mass of water = 250 g Molar mass of water (H$_2$O) = 18 g/mol Moles of water ($n_{\text{water}}$) = $\frac{\text{Mass}}{\text{Molar Mass}} = \frac{250 \text{ g}}{18 \text{ g/mol}} \approx 13.888... \text{ mol}$

2. Calculate Molality (m):

Molality is defined as the number of moles of solute per kilogram of solvent.

• Mass of solvent (water) in kg = $250 \text{ g} \times \frac{1 \text{ kg}}{1000 \text{ g}} = 0.250 \text{ kg}$
• Molality (m) = $\frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} = \frac{0.05 \text{ mol}}{0.250 \text{ kg}} = 0.2 \text{ mol/kg} = 0.2 \text{ m}$

3. Calculate Mole fraction of urea ($X_{\text{urea}}$):

Mole fraction of a component is the ratio of the moles of that component to the total moles of all components in the solution.

• Total moles ($n_{\text{total}}$) = Moles of urea + Moles of water $n_{\text{total}} = 0.05 \text{ mol} + 13.888... \text{ mol} = 13.938... \text{ mol}$
• Mole fraction of urea ($X_{\text{urea}}$) = $\frac{\text{Moles of urea}}{\text{Total moles}} = \frac{0.05 \text{ mol}}{13.938... \text{ mol}} \approx 0.003587$
• Rounding to five decimal places, $X_{\text{urea}} \approx 0.00359$

Comparing the calculated values with the given options, the option "0-2 m and 0-00359" matches our results (assuming the hyphen '0-' represents a decimal point).