Question

how many kilograms of solvent must .71 moles of Kl be dissolved in to produce a 1.93 m solution?

Answer 1

From equation for molar concentration $c=\frac{n}{V}$ we can found volume of 1.93M solution containing 71 moles of KI:

$V=\frac{n}{c}=\frac{71 mol}{1.93\frac{mol}{L}}=36.8L$

Taking into account that the density of 1.93M solution is about 1.22 kg/L (tabulated data), mass of the solution can be found:

$m(solution)=d \cdot V=1.22\frac{kg}{L} \cdot 36.8L = 44.9 kg$

Thus the mass of solvent can be calculated:

$m(solvent)=m(solution)-m(KI)=m(solution)-n(KI) \cdot M(KI)=44.9kg-71mol \cdot 166 \times 10^{-3} \frac{kg}{mol}= 44.9kg-11.8kg=33.1kg$