Answer
a. 20. g of $LiNO_3$ solution
b. 400. mL of $KOH$ solution.
c. 20. mL of $HCO_2H$ (formic acid) solution.
Work Step by Step
a. $25\%$ (m/m) $ LiNO_3 $ solution : $$ 25 \space g \space LiNO_3 = 100 \space g \space solution$$
$$5.0 \space g \space LiNO_3 \times \frac{100 \space g \space solution}{25 \space g \space LiNO_3} = 20. \space g \space solution$$
b. $10.0\%$ (m/v) $ KOH $ solution : $$ 10.0 \space g \space KOH = 100 \space mL \space solution$$
$$40.0 \space g \space KOH \times \frac{100 \space mL \space solution}{10.0 \space g \space KOH} = 400. \space mL \space solution$$
c. $10.0\%$ (v/v) $ formic \space acid $ solution : $$ 10.0 \space mL \space formic \space acid = 100 \space mL \space solution$$
$$2.0 \space mL \space formic \space acid \times \frac{100 \space mL \space solution}{10.0 \space mL \space formic \space acid} = 20. \space mL \space solution$$
