Answer
a. $2.5 \space g \space KCl $
b. $50. \space g \space NH_4Cl $
c. $25.0 \space mL \space acetic \space acid $
Work Step by Step
a.
$5.0\%$ (m/m) KCl solution: $$ 5.0 \space g \space KCl = 100 \space g \space solution$$
$$50. \space g \space solution \times \frac{5.0 \space g \space KCl}{100 \space g \space solution} = 2.5 \space g \space KCl $$
b.
$4.0\%$ (m/v) $ NH_4Cl $ solution : $$ 4.0 \space g \space NH_4Cl = 100 \space mL \space solution$$
$$1250 \space mL \space solution \times \frac{4.0 \space g \space NH_4Cl}{100 \space mL \space solution} = 50. \space g \space NH_4Cl $$
c.
$10.0\%$ (v/v) $ acetic \space acid $ solution : $$ 10.0 \space mL \space acetic \space acid = 100 \space mL \space solution$$
$$250. \space mL \space solution \times \frac{10.0 \space mL \space acetic \space acid}{100 \space mL \space solution} = 25.0 \space mL \space acetic \space acid $$
