Answer
There are necessary 60.0 g of $LiNO_3$ to prepare 150. g of that solution.
Work Step by Step
1. Identify the conversion factors:
- Since the $LiNO_3$ solution is 40.0% (m/m):
40.0 g (Solute) = 100. g (Solution)
So: $\frac{40.0 \space g \space (Solute)}{100. \space g \space (Solution)}$ and $\frac{100. \space g \space (Solution)}{40.0 \space g \space (Solute)}$
2. Calculate the mass of solute needed to prepare 150. g of that solution:
$150. \space g \space (Solution) \times \frac{40.0 \space g \space (Solute)}{100. \space g \space (Solution)} = 60.0 \space g \space (Solute)$
