Q. 9.49 - An Introduction to General, Organic, and Biological Chemistry

Question

For each of the following solutions, calculate the:

a. liter of a $2.00 \mathrm{M} \mathrm{KBr}$ solution to obtain $3.00 \mathrm{moles} \mathrm{of} \mathrm{KBr}$.

b. liter of a $1.50 \mathrm{M} \mathrm{NaCl}$ solution to obtain $15.0 \mathrm{moles} \mathrm{of} \mathrm{NaCl}$.

c. milliliter of a $0.800 \mathrm{M} \mathrm{Ca}{\left({\mathrm{NO}}_3\right)}_2$ solution to obtain $0.0500 \mathrm{mole} \mathrm{of} \mathrm{Ca}{\left({\mathrm{Na}}_3\right)}_2$.

Step-by-Step Solution

Answer

a. The liter of $2.00 \mathrm{M} \mathrm{KBr}$  solution is $1.5 \mathrm{L}$.

b. The liter of $1.50 \mathrm{M} \mathrm{NaCl}$ solution is $10.0 \mathrm{L}$.

c. The milliliter of $0.800 \mathrm{M} \mathrm{Ca}{\left({\mathrm{NO}}_3\right)}_2$ solution is $62.5 \mathrm{mL}$.

Part (a) step 1: Given Information

We need to calculate the liter of a $2.00 \mathrm{M} \mathrm{KBr}$ solution to obtain $3.00 \mathrm{moles} \mathrm{of} \mathrm{KBr}$. 

Part (a) step 2: Explanation

Consider:

$$\begin{gathered} \mathrm{n}\left(\mathrm{KBr}\right)=3.00 \mathrm{moles} \\ \mathrm{M}=2.00 \mathrm{M}=2.00 \raisebox{1ex}{$\mathrm{mol}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right. \end{gathered}$$

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

Now,
$$\begin{gathered} \mathrm{V}\left(\mathrm{KBr}\right)=\frac{\mathrm{n}\left(\mathrm{KBr}\right)}{\mathrm{M}} \\ =\frac{3.00 \mathrm{mole}}{2.00{\displaystyle \raisebox{1ex}{$ \mathrm{mol}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right.}} \\ =1.5 \mathrm{L} \end{gathered}$$

Part (b) step 1: Given Information

We need to calculate the liter of a $1.50 \mathrm{M} \mathrm{NaCl}$solution to obtain $15.0 \mathrm{moles} \mathrm{of} \mathrm{NaCl}$.

Part (b) step 2: Explanation

Consider:

$$\begin{gathered} \mathrm{n}\left(NaCl\right)=15.0 \mathrm{moles} \\ \mathrm{M}=1.5 \mathrm{M}=1.5 \raisebox{1ex}{$\mathrm{mol}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right. \end{gathered}$$

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

Now,

$$\begin{gathered} \mathrm{V}\left(\mathrm{NaCl}\right)=\frac{\mathrm{n}\left(\mathrm{NaCl}\right)}{\mathrm{M}} \\ =\frac{15.0 \mathrm{mole}}{{\displaystyle 1.5\raisebox{1ex}{$ \mathrm{mol}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right.}} \\ =10.0 \mathrm{L} \end{gathered}$$

Part (c) step 1: Given Information

We need to calculate the milliliter of a  $0.800 \mathrm{M} \mathrm{Ca}{\left({\mathrm{NO}}_3\right)}_2$  solution to obtain $0.0500 \mathrm{mole} \mathrm{of} \mathrm{Ca}{\left({\mathrm{Na}}_3\right)}_2$.

Part (c) step 2: Explanation

Considering

$$\begin{gathered} \mathrm{n}\left(Ca{\left(N{a}_3\right)}_2\right)=0.0500 \mathrm{moles} \\ \mathrm{M}=0.800 \mathrm{M}=0.800 \raisebox{1ex}{$\mathrm{mol}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right. \end{gathered}$$

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

V = $\frac{0.05}{0.8}=62.5mL$