Question

What is the mole fraction of calcium chloride in 3.35 m CaCl2(aq)? The molar mass of CaCl2 is 111.0 g/mol and the molar mass of water is 18.02 g/mol.

Answer 1

Answer: The mole fraction of calcium chloride and water in the solution is 0.057 and 0.943 respectively

Explanation:

We are given:

Molality of calcium chloride = 3.35 m

This means that 3.35 moles of calcium chloride are present in 1 kg or 1000 g of water

To calculate the number of moles, we use the equation:

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

Given mass of water = 1000 g

Molar mass of water = 18.02 g/mol

Putting values in above equation, we get:

$\text{Moles of water}=\frac{1000g}{18.02g/mol}=55.49mol$

Total moles of solution = [3.35 + 55.49] = 58.84 moles

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

• For calcium chloride:

$\chi_{CaCl_2}=\frac{n_{CaCl_2}}{n_{CaCl_2}+n_{H_2O}}\\\\\chi_{CaCl_2}=\frac{3.35}{58.84}=0.057$

• For water:

$\chi_{H_2O}=\frac{n_{H_2O}}{n_{CaCl_2}+n_{H_2O}}\\\\\chi_{H_2O}=\frac{55.49}{58.84}=0.943$

Hence, the mole fraction of calcium chloride and water in the solution is 0.057 and 0.943 respectively

Answer 2

Answer:

49.3% water

Explanation: