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 $0.0500\mathrm{mole}\mathrm{of}\mathrm{Ca}{\left({\mathrm{Na}}_3\right)}_2$.

Short 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 $62.5\mathrm{mL}$.

Step-by-step solution

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 =$\frac{0.05}{0.8}=62.5mL$