# Question #ddc52

(a) 10 L soln × $\left(1.18 \text{ kg soln")/(1" L soln") × (24" kg NaCl")/(100" kg soln}\right)$ = 2.8 kg NaCl
(b) 10 kg NaCl × $\left(100 \text{ kg soln")/(24" kg NaCl") × (1" L soln")/(1.18" kg soln}\right)$ = 35 L soln