A $10 L$ sample of hard water contains $40 mg L^{- 1}$ of $Ca X^{2 +}$ and $12 mg L^{- 1}$ of $Mg X^{2 +}$ (Ca = 40, Mg = 24). Both ions are precipitated completely by sodium stearate (molar mass $306$ g/mol) before any lather forms. The mass of sodium stearate consumed is

Answer: D

Moles of

Ca X^{2 +}

0.400/40 = 0.0100

mol and of

Mg X^{2 +}

0.120/24 = 0.00500

mol, total

0.0150

mol of divalent ions. Each needs 2 stearate ions, so soap =

0.0300

mol. Mass =

0.0300 \times 306 = 9.18 g