Question

Mole fraction of A in H2O is 0.2. The molality of A in H2O is:

Answer 1

Answer: The molality of the A in water is 13.88 mol/kg.

Explanation:

We are given:

Mole fraction of A is 0.2, which means that 0.2 moles of A is present in 1 mole of solution.

Mole fraction of B $(H_2O)=1-\chi_A=1-0.2=0.8$

This means that 0.8 moles of B is present in 1 mole of solution.

To calculate the mass of water, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of water = 0.8 mol

Molar mass of water = 18 g/mol

Putting values in above equation, we get:

$0.8=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=14.4g$

To calculate the molality, we use the equation:

$\text{Molality}=\frac{\text{Moles of solute}}{\text{Mass of solvent (in kg)}}$

We are given:

Moles of solute (A) = 0.2 mol

Mass of solvent (water) = 14.4 g = 0.0144 kg     (Conversion factor: 1 kg = 1000g)

Putting values in above equation, we get:

$\text{Molality}=\frac{0.2mol}{0.0144kg}=13.88mol/kg$

Hence, the molality of the A in water is 13.88 mol/kg.

Answer 2

Answer:

the molality of the A in water is 13.88 mol/kg.

Explanation:

Mole fraction of A is 0.2, which means that 0.2 moles of A is present in 1 mole of solution.

Mole fraction of B  

This means that 0.8 moles of B is present in 1 mole of solution.

To calculate the mass of water, we use the equation:

Moles of water = 0.8 mol

Molar mass of water = 18 g/mol

Putting values in above equation, we get:

To calculate the molality, we use the equation:

We are given:

Moles of solute (A) = 0.2 mol

Mass of solvent (water) = 14.4 g = 0.0144 kg     (Conversion factor: 1 kg = 1000g)

Putting values in above equation, we get:

Hence, the molality of the A in water is 13.88 mol/kg.