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

Short 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$5.47\mathrm{g}$.

Step-by-step solution

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 =$\frac{\text{mass}}{\text{molarmass}}$

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

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 =$\frac{\text{mass}}{\text{molarmass}}$

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

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 =$\frac{\text{mass}}{\text{molarmass}}$

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