Question

Calculate the mole fraction of 5.65g Calcium Chloride with molar mass of 111g/mole placed in 15.5g of H2O with molar mass of 18.0g/mole. Equation please

Answer 1

n(CaCl₂) = 5.65 / 111 = 0.51 mol

n(H₂O) = 15.5 / 18 = 0.86 mol

$\omega=\dfrac{n(CaCl_2)}{n(CaCl_2)+n(H_2O)} = \dfrac{0.51}{0.86+0.51} = 0.372$