Answer
a. 60.0 g $LiNO_3$
b. 9.0 g $KOH$
c. 25.0 mL acetic acid
Work Step by Step
a.
$40.0\%$ (m/m) $ LiNO_3 $ solution : $$ 40.0 \space g \space LiNO_3 = 100 \space g \space solution$$
$$150. \space g \space solution \times \frac{40.0 \space g \space LiNO_3}{100 \space g \space solution} = 60.0 \space g \space LiNO_3 $$
b.
$2.0\%$ (m/v) $ KOH $ solution : $$ 2.0 \space g \space KOH = 100 \space mL \space solution$$
$$450 \space mL \space solution \times \frac{2.0 \space g \space KOH}{100 \space mL \space solution} = 9.0 \space g \space KOH $$
c.
$10.0\%$ (v/v) $ acetic \space acid $ solution : $$ 10.0 \space mL \space acetic \space acid = 100 \space mL \space solution$$
$$250. \space mL \space solution \times \frac{10.0 \space mL \space acetic \space acid}{100 \space mL \space solution} = 25.0 \space mL \space acetic \space acid $$
