Chemistry: What did this lesson make you curious about? What other questions do you have about hair?

of the stock solution would be required. Explanation:Assume that a solution of volume contains a solute with a concentration of . The quantity of that solute in this solution would be: .For the solution that needs to be prepared, . The volume of this solution is . Calculate the quantity of the solute (magnesium chloride) in the required solution:.Rearrange the equation to find an expression of volume , given the concentration and quantity of the solute:.Concentration of the solute in the stock...

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hernandez09297

23.01.2023

Chemistry

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

Explanation:

Assume that a solution of volume contains a solute with a concentration of . The quantity 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 , given the concentration and quantity 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:

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

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