Question

Mole fraction of $\mathrm{A}$ in ${\mathrm{H}}_2\mathrm{O}$ is $0.2$ the molality of $\mathrm{A}$ in ${\mathrm{H}}_2\mathrm{O}$ is:

Answer 1

Step 1: Find mole fraction of B.

• We are given that Mole fraction of $\mathrm{A}$ is $0.2$, which means that $0.2$ moles of $\mathrm{A}$ is present in one mole of solution.
• Mole fraction of $\mathrm{B}$ is:
  $1-X_A=1-0.2=0.8$
• This means that $0.8$ moles of $\mathrm{B}$ is present in one mole of solution.

Step 2: Calculate mass of water

• To calculate the mass of water, we use the equation:
  $\mathrm{Number}\mathrm{of}\mathrm{moles}=\frac{\mathrm{Given}\mathrm{mass}}{\mathrm{Molar}\mathrm{mass}}$
• Moles of water $=0.8$
• The molar mass of water $18\mathrm{g}{\mathrm{mol}}^{-1}$
• Putting values in the above equation, we get:
  $0.8=\frac{\text{Mass of water}}{18\mathrm{g}{\mathrm{mol}}^{-1}}$
• So, the Mass of water $14.4\mathrm{g}$

Step 3: Calculate molality of $\mathrm{A}$ in Water

• To calculate the molality, we use the equation:
  $\text{Molality}=\frac{\text{Moles of solute}}{\text{Mass of solvent}\left(\mathrm{kg}\right)}$
• We are given:
  Moles of solute $=0.2\mathrm{mol}$
  Mass of solvent (water) $=14.4\mathrm{g}=0.0144\mathrm{kg}$
• Putting values in above equation, we get:

$\mathrm{Molality}=\frac{0.2\text{mole}}{0.0144\mathrm{kg}}=13.88\mathrm{mol}{\mathrm{kg}}^{-1}$

Hence, the molality of the $\mathrm{A}$ in water is $13.88\mathrm{mol}{\mathrm{kg}}^{-1}$