calculate the mass of impure KOH needed to make up 1.20 L of 0.60 mol/L KOH(aq)? Assume the impure KOH is 84% KOH by mass and 16% water.

Answer 1

try this;

n = Mv = 0.6x1.2 = 0.72moles

so, 0.72moles is the total mole i.e. KOH and water. Since the impure KOH is 84%, then the mole for KOH is;

0.84 x 0.72 = 0.605moles

so, m = nMr = 0.605x(molar mass of KOH).