Q. 9.48 - An Introduction to General, Organic, and Biological Chemistry

Question

Calculate the grams of solute needed to prepare each of the following: 

a. $2.00 \mathrm{L} \mathrm{of} \mathrm{a} 6.00 \mathrm{M} \mathrm{NaOH}$solution

b. $5.00 \mathrm{L} \mathrm{of} \mathrm{a} 0.100 \mathrm{M} {\mathrm{CaCl}}_2$ solution

c. $175 \mathrm{mL} \mathrm{of} \mathrm{a} 3.00 \mathrm{M} {\mathrm{NaNO}}_3$ solution 

Step-by-Step Solution

Answer

a. The grams of  $2.00 \mathrm{L} \mathrm{of} \mathrm{a} 6.00 \mathrm{M} \mathrm{NaOH}$is $48 \mathrm{g}$.

b. The grams of  $5.00 \mathrm{L} \mathrm{of} \mathrm{a} 0.100 \mathrm{M} {\mathrm{CaCl}}_2$ is $55.45 \mathrm{g}$.

c. The grams of  $175 \mathrm{mL} \mathrm{of} \mathrm{a} 3.00 \mathrm{M} {\mathrm{NaNO}}_3$ is $44.63 \mathrm{g}$.

Part (a) step 1: Given Information

We need to calculate the grams of $2.00 \mathrm{L} \mathrm{of} \mathrm{a} 6.00 \mathrm{M} \mathrm{NaOH}$ the solution.

Part (a) step 2: Explanation

Considering,

$$\begin{gathered} \mathrm{V}\left(\mathrm{NaOH}\right)=2.00 \mathrm{L} \\ \mathrm{M}=6.00 \mathrm{M}=6.00 \raisebox{1ex}{$\mathrm{mole}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right. \end{gathered}$$

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

Now,

$$\begin{gathered} \mathrm{n}\left(\mathrm{NaOH}\right)=\mathrm{M} \times \mathrm{V}\left(\mathrm{NaOH}\right) \\ =6.00 \raisebox{1ex}{$\mathrm{mole}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right. \times 2.0 \mathrm{L} \\ =12.0 \mathrm{mole} \end{gathered}$$

Number of moles$\mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}}$

$\mathrm{m}=\mathrm{n}\times \mathrm{M}$

So,

$$\begin{gathered} \mathrm{m}\left(\mathrm{NaOH}\right)=\mathrm{n}\left(\mathrm{NaOH}\right)\times \mathrm{M}\left(\mathrm{NaOH}\right) \\ =12.0 \mathrm{mole} \times \left(23+1+6\right) \raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right. \\ =48 \mathrm{g} \end{gathered}$$

Part (b) step 1: Given Information

We need to calculate the grams of $5.00 \mathrm{L} \mathrm{of} \mathrm{a} 0.100 \mathrm{M} {\mathrm{CaCl}}_2$the solution. 

Part (b) step 2: Explanation

Considering

$$\begin{gathered} \mathrm{V}\left({\mathrm{CaCl}}_2\right)=5.00 \mathrm{L} \\ \mathrm{M}=0.100 \mathrm{M}=0.100 \raisebox{1ex}{$\mathrm{mole}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right. \end{gathered}$$

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

Now,

$$\begin{gathered} \mathrm{n}\left({\mathrm{CaCl}}_2\right)=\mathrm{M} \times \mathrm{V}\left({\mathrm{CaCl}}_2\right) \\ =0.100 \raisebox{1ex}{$\mathrm{mole}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right. \times 5.0 \mathrm{L} \\ =0.50 \mathrm{mole} \end{gathered}$$

Number of moles$\mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}}$

$\mathrm{m}=\mathrm{n}\times \mathrm{M}$

So, 

$$\begin{gathered} \mathrm{m}\left({\mathrm{CaCl}}_2\right)=\mathrm{n}\left({\mathrm{CaCl}}_2\right)\times \mathrm{M}\left({\mathrm{CaCl}}_2\right) \\ =0.50 \mathrm{mole} \times \left(2\times 35.45+40\right) \raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right. \\ =55.45 \mathrm{g} \end{gathered}$$

Part (c) step 1: Given Information

We need to calculate the grams of $175 \mathrm{mL} \mathrm{of} \mathrm{a} 3.00 \mathrm{M} {\mathrm{NaNO}}_3$ the solution. 

Part (c) step 2: Simplify

Considering

V = $175mL=0.175L$

M = $3M$

$\mathrm{n}=\mathrm{M}\times \mathrm{V}$

Now,

$$\begin{gathered} \mathrm{n}\left({\mathrm{NaNO}}_3\right)=\mathrm{M} \times \mathrm{V}\left({\mathrm{NaNO}}_3\right) \\ =3.000 \raisebox{1ex}{$\mathrm{mole}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right. \times 0.175 \mathrm{L} \\ =0.525 \mathrm{mole} \end{gathered}$$

The number of moles$\mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}}$

$\mathrm{m}=\mathrm{n}\times \mathrm{M}$

So, 

$$\begin{gathered} \mathrm{m}\left({\mathrm{NaNO}}_3\right)=\mathrm{n}\left({\mathrm{NaNO}}_3\right)\times \mathrm{M}\left({\mathrm{NaNO}}_3\right) \\ =0.525 \mathrm{mole} \times \left(23+14+3+16\right) \raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right. \\ =44.63 \mathrm{g} \end{gathered}$$