Question

For each of the following solutions, calculate the:

a. liters of a $4.00\mathrm{M}\mathrm{KCl}$solution to obtain $0.100\mathrm{mole}\mathrm{of}\mathrm{KCl}$.

b. liters of a $6.00\mathrm{M}\mathrm{HCl}$solution to obtain $5.00\mathrm{moles}\mathrm{of}\mathrm{HCl}$.

c. milliliters of a $2.50\mathrm{M}{\mathrm{K}}_2{\mathrm{SO}}_4$solution to obtain$1.20\mathrm{moles}\mathrm{of}{\mathrm{K}}_2{\mathrm{SO}}_4$.

Short answer

a. The liter of $4.00\mathrm{M}\mathrm{KCl}$solution is $0.025\mathrm{L}$.

b. The liter of $6.00\mathrm{M}\mathrm{HCl}$ solution is $8.33\mathrm{L}$.

c. The milliliter of $2.50\mathrm{M}{\mathrm{K}}_2{\mathrm{SO}}_4$solution is $480\mathrm{mL}$.

Step-by-step solution

Part (a) step 1: Given Information

We need to calculate the liter of a$4.00\mathrm{M}\mathrm{KCl}$ solution to obtain$0.100\mathrm{mole}\mathrm{of}\mathrm{KCl}$.

Part (a) step 2: Explanation

Consider:

$$\begin{gathered} \mathrm{n}\left(K\mathrm{Cl}\right)=0.100\mathrm{moles} \\ \mathrm{M}=4.00\mathrm{M}=4.00\raisebox{1ex}{$\mathrm{mol}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right. \end{gathered}$$

We know$\mathrm{V}=\frac{\mathrm{n}}{\mathrm{M}}$

Now,

$$\begin{gathered} \mathrm{V}\left(\mathrm{KCl}\right)=\frac{\mathrm{n}\left(\mathrm{KCl}\right)}{\mathrm{M}} \\ =\frac{0.100\mathrm{mole}}{4.0{\displaystyle \raisebox{1ex}{$\mathrm{mol}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right.}} \\ =0.025\mathrm{L} \end{gathered}$$

Part (b) step 1: Given Information

We need to calculate the liter of a$6.00\mathrm{M}\mathrm{HCl}$ solution to obtain$5.00\mathrm{moles}\mathrm{of}\mathrm{HCl}$.

Part (b) step 2: Explanation

Consider:

$$\begin{gathered} \mathrm{n}\left(\mathrm{HCl}\right)=5.00\mathrm{moles} \\ \mathrm{M}=6.00\mathrm{M}=6.00\raisebox{1ex}{$\mathrm{mol}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right. \end{gathered}$$

We know$\mathrm{V}=\frac{\mathrm{n}}{\mathrm{M}}$

Now,

$$\begin{gathered} \mathrm{V}\left(\mathrm{HCl}\right)=\frac{\mathrm{n}\left(\mathrm{HCl}\right)}{\mathrm{M}} \\ =\frac{5.00\mathrm{mole}}{6.00{\displaystyle \raisebox{1ex}{$\mathrm{mol}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right.}} \\ =8.33\mathrm{L} \end{gathered}$$

Part (c) step 1: Given Information

We need to calculate the milliliter of a $2.50\mathrm{M}{\mathrm{K}}_2{\mathrm{SO}}_4$solution to obtain$1.20\mathrm{moles}\mathrm{of}{\mathrm{K}}_2{\mathrm{SO}}_4$.

Part (c) step 2: Explanation

Consider:

$$\begin{gathered} \mathrm{n}\left({\mathrm{K}}_2{\mathrm{SO}}_4\right)=1.20\mathrm{moles} \\ \mathrm{M}=2.50\mathrm{M}=2.50\raisebox{1ex}{$\mathrm{mol}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right. \end{gathered}$$

We know$\mathrm{V}=\frac{\mathrm{n}}{\mathrm{M}}$

Now,

$$\begin{gathered} \mathrm{V}\left({\mathrm{K}}_2{\mathrm{SO}}_4\right)=\frac{\mathrm{n}\left({\mathrm{K}}_2{\mathrm{SO}}_4\right)}{\mathrm{M}} \\ =\frac{1.20\mathrm{mole}}{2.50{\displaystyle \raisebox{1ex}{$\mathrm{mol}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right.}} \\ =0.48\mathrm{L} \end{gathered}$$

Now, convert liter into milliliter:

$0.48\mathrm{L}\times \frac{1000\mathrm{L}}{1\mathrm{L}}=480\mathrm{mL}$