Answer
There are necessary 50. g of $NH_4Cl$ to prepare 1250 mL of that solution.
Work Step by Step
1. Identify the conversion factors:
- Since the $NH_4Cl$ solution is 4.0% (m/v):
4.0 g (Solute) = 100 mL (Solution)
So: $\frac{4.0 \space g \space (Solute)}{100 \space mL \space (Solution)}$ and $\frac{100 \space mL \space (Solution)}{4.0 \space g \space (Solute)}$
2. Calculate the mass of solute needed to prepare 1250 mL of that solution:
$1250 \space mL \space (Solution) \times \frac{4.0 \space g \space (Solute)}{100 \space mL \space (Solution)} = 50. \space g \space (Solute)$