Question

You are given a stock solution of 500.0 mL of 1.00M magnesium chloride solution. Calculate the volume of the stock solution you would need to use to prepare 250.0 mL of 0.20 M solution.

Answer 1

Answer:

$50\; \rm mL$ of the stock solution would be required.

Explanation:

Assume that a solution of volume $V$ contains a solute with a concentration of $c$. The quantity $n$ of that solute in this solution would be:

$n = c \cdot V$.

For the solution that needs to be prepared, $c = 0.20\; \rm M = 0.20\; \rm mol \cdot L^{-1}$. The volume of this solution is $V = 250.0\; \rm mL$. Calculate the quantity of the solute (magnesium chloride) in the required solution:

$\begin{aligned}n &= c \cdot V \\ &= 0.20\; \rm mol \cdot L^{-1} \times 250.0\; \rm mL \\ &= 0.20\; \rm mol \cdot L^{-1} \times 0.2500\; \rm L \\ &= 0.050\; \rm mol\end{aligned}$.

Rearrange the equation $n = c \cdot V$ to find an expression of volume $V$, given the concentration $c$ and quantity $n$ of the solute:

$\displaystyle V= \frac{n}{c}$.

Concentration of the solute in the stock solution: $c(\text{stock}) = 1.00\; \rm M = 1.00\; \rm mol \cdot L^{-1}$.

Quantity of the solute required: $n = 0.050\; \rm mol$.

Calculate the volume of the stock solution that would contain the required $n = 0.050\; \rm mol$ of the magnesium chloride solute:

$\begin{aligned}& V(\text{stock}) \\ &= \frac{n}{c(\text{stock})} \\ &= \frac{0.050\; \rm mol}{1.00\; \rm mol \cdot L^{-1}} \\ &= 0.050\; \rm L \\ &= 50\; \rm mL\end{aligned}$.