Answer
20. g of that solution contains 5.0 g of $LiNO_3$.
Work Step by Step
1. Identify the conversion factors:
- Since the $LiNO_3$ solution is 25% (m/m):
25 g (Solute) = 100. g (Solution)
So: $\frac{25 \space g \space (Solute)}{100. \space g \space (Solution)}$ and $\frac{100. \space g \space (Solution)}{25 \space g \space (Solute)}$
2. Calculate the mass of solution that contains 5.0 g of that solute $(LiNO_3)$:
$5.0 \space g \space (Solute) \times \frac{100. \space g \space (Solution)}{25 \space g \space (Solute)} = 20. \space g \space (Solution)$
