Question

40.0 grams of Calcium Carbonate compound (CaCO3) with molar mass of 100g/mole, in 

order to make an aqueous solution that has a mole fraction of solute of 0.200? How much

H2O is needed to achieved the desired result?

Answer 1

Mass of CaCO₃ = 40 g

Molar mass of CaCO₃ = 100 g/mol

Moles of CaCO₃(n₁) = $\dfrac{40}{100}=0.4\space mol$

Mole fraction of solute CaCO₃(x₁) = 0.200

Moles of solvent H₂O = n₂

$\dfrac{n_1}{n_1+n_2}=0.200\\\Rightarrow\dfrac{0.4}{0.4+n_2}=0.200$

$\Rightarrow n_2=1.6\space mol$

Moles of H₂O = 1.6 mol

Mass of H₂O needed = $1.6\times18=28.8\space g$