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

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

Step-by-step 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) * 250.0\; \rm mL \\ &= 0.20\; \rm mol \cdot L^(-1) * 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= (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})} \\ &= (0.050\; \rm mol)/(1.00\; \rm mol \cdot L^(-1)) \\ &= 0.050\; \rm L \\ &= 50\; \rm mL\end{aligned}`.