Q. 9.47 - An Introduction to General, Organic, and Biological Chemistry

Question

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

a. $2.00 \mathrm{L} \mathrm{of} \mathrm{a} 1.50 \mathrm{M} \mathrm{NaOH}$solution

b. $4.00 \mathrm{L} \mathrm{of} \mathrm{a} 0.200 \mathrm{M} \mathrm{KCl}$ solution

c. $25.0 \mathrm{L} \mathrm{of} \mathrm{a} 6.00 \mathrm{M} \mathrm{HCl}$  solution

Step-by-Step Solution

Answer

a.  The grams of $2.00 \mathrm{L} \mathrm{of} \mathrm{a} 1.50 \mathrm{M} \mathrm{NaOH}$ is $120 \mathrm{g}$.

b.  The grams of  $4.00 \mathrm{L} \mathrm{of} \mathrm{a} 0.200 \mathrm{M} \mathrm{KCl}$ is $74.55 \mathrm{g}$.

c.   The grams of $25.0 \mathrm{L} \mathrm{of} \mathrm{a} 6.00 \mathrm{M} \mathrm{HCl}$ is $5.47 \mathrm{g}$.

Part (a) step 1: Given Information

We need to calculate the grams of $2.00 \mathrm{L} \mathrm{of} \mathrm{a} 1.50 \mathrm{M} \mathrm{NaOH}$solution.

Part (a) step 2: Explanation

Considering

$$\begin{gathered} \mathrm{V}\left(\mathrm{NaOH}\right)=2.00 \mathrm{L} \\ \mathrm{M}=1.5 \mathrm{M}= 1.5 \raisebox{1ex}{$\mathrm{mole}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right. \end{gathered}$$

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

Now, n = $1.5\times 2=3 \text{mole}$

Number of moles = $\frac{\text{mass}}{\text{molar mass}}$

$$\begin{gathered} \mathrm{m}\left(\mathrm{NaOH}\right)=\mathrm{n}\left(\mathrm{NaOH}\right)\times \mathrm{M}\left(\mathrm{NaOH}\right) \\ =3.0 \mathrm{mole} \times \left(23+1+6\right)\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right. \\ =120 \mathrm{g} \end{gathered}$$

Part (b) step 1: Given Information

We need to calculate the grams of $4.00 \mathrm{L} \mathrm{of} \mathrm{a} 0.200 \mathrm{M} \mathrm{KCl}$ solution.

Part (b) step 2: Explanation

Considering,

$$\begin{gathered} \mathrm{V}\left(\mathrm{KCl}\right)=4.00 \mathrm{L} \\ \mathrm{M}=0.200 \mathrm{M}=0.200 \raisebox{1ex}{$\mathrm{mole}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right. \end{gathered}$$

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

Now,

$$\begin{gathered} \mathrm{n}\left(KCl\right)=\mathrm{M} \times \mathrm{V}\left(KCl\right) \\ =0.200 \raisebox{1ex}{$\mathrm{mole}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right. \times 4.00 \mathrm{L} \\ =0.800 \mathrm{mole} \end{gathered}$$

The number of moles = $\frac{\text{mass}}{\text{molar mass}}$

$$\begin{gathered} \mathrm{m}\left(\mathrm{KCl}\right)=\mathrm{n}\left(\mathrm{KCl}\right)\times \mathrm{M}\left(\mathrm{KCl}\right) \\ =0.800 \mathrm{mole} \times \left(39.10+35.45\right) \raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right. \\ =74.55 \mathrm{g} \end{gathered}$$

Part (c) step 1: Given Information

We need to calculate the grams of $25.0 \mathrm{L} \mathrm{of} \mathrm{a} 6.00 \mathrm{M} \mathrm{HCl}$ solution.

Part (c) step 2: Explanation

Considering,

V = $25L$HCL

M = $6\text{M}$

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

Now,

$$\begin{gathered} \mathrm{n}\left(\mathrm{HCl}\right)=\mathrm{M} \times \mathrm{V}\left(\mathrm{HCl}\right) \\ =6.00 \raisebox{1ex}{$\mathrm{mole}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right. \times 0.25 \mathrm{L} \\ =0.15 \mathrm{mole} \end{gathered}$$

The number of moles = $\frac{\text{mass}}{\text{molar mass}}$

$$\begin{gathered} \mathrm{m}\left(\mathrm{HCl}\right)=\mathrm{n}\left(\mathrm{HCl}\right)\times \mathrm{M}\left(\mathrm{HCl}\right) \\ =0.15 \mathrm{mole} \times 36.45 \raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right. \\ =5.47 \mathrm{g} \end{gathered}$$