Q. 9.50 - An Introduction to General, Organic, and Biological Chemistry

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$1.20 \mathrm{moles} \mathrm{of} {\mathrm{K}}_2{\mathrm{SO}}_4$.

Step-by-Step Solution

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 $480 \mathrm{mL}$.

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:

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