Question

Calculate each of the following quantities:

(a) Volume in milliliters of 2.26 M potassium hydroxide that contains 8.42 g of solute

(b) Number of Cu2+ions in 52 L of 2.3 M copper(II) chloride

(c) Molarity of 275 mL of solution containing 135 mmol of glucose

Short answer

Answer

1. Volume of 2.26 M potassium hydroxide that contains 8.42 g of soluteis 66.37 mL.

2. Number of Cu2+ions in 52 L of 2.3 M copper(II) chloride is 7.2× 1025.

3. Molarity of 275 mL of solution containing 135 mmol of glucose is 0.49 M.

Step-by-step solution

Calculation of Volume in milliliters of potassium hydroxide solution

First of all, we need to calculate the moles of the solute

$\mathit{M}\mathit{o}\mathit{l}\mathit{e}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{S}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{e}\mathbf{=}\frac{\mathbf{M}\mathbf{a}\mathbf{s}\mathbf{s}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{s}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{t}\mathbf{e}}{\mathbf{M}\mathbf{o}\mathbf{l}\mathbf{a}\mathbf{r}\mathbf{}\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}}\mathbf{=}\frac{\mathbf{8}\mathbf{.}\mathbf{42}}{\mathbf{56}\mathbf{.}\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{15}$

Molarity is given by the formula,

$\mathit{M}\mathit{o}\mathit{l}\mathit{e}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{S}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{e}\mathbf{=}\frac{\mathbf{M}\mathbf{a}\mathbf{s}\mathbf{s}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{s}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{t}\mathbf{e}}{\mathbf{V}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{m}\mathbf{e}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{s}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{r}\mathbf{i}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}\mathbf{}\left(mL\right)}\mathbf{\times }\mathbf{1000}$

Using the given data, we can say

$$\begin{gathered} \mathbf{2}\mathbf{.}\mathbf{26}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{15}}{\mathbf{Volume}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solurition}\mathbf{}\mathbf{\left(}\mathbf{mL}\mathbf{\right)}}\mathbf{\times }\mathbf{1000} \\ \mathit{V}\mathit{o}\mathit{l}\mathit{u}\mathit{m}\mathit{e}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\left(mL\right)\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{15}\mathbf{\times }\mathbf{1000}}{\mathbf{2}\mathbf{.}\mathbf{26}}\mathbf{=}\mathbf{66}\mathbf{.}\mathbf{37}\mathbf{}\mathit{m}\mathit{L} \end{gathered}$$

Hence, volume of the given solution is 66.37 mL.

Calculation of Number of Cu2+ ions in copper(II) chloride

For calculating number of ions first let us calculated the moles of CaCl2in the solution.

$$\begin{gathered} \mathit{M}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathit{i}\mathit{t}\mathit{y}\mathbf{=}\frac{\mathbf{M}\mathbf{o}\mathbf{l}\mathbf{e}\mathbf{s}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{s}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{t}\mathbf{e}}{\mathbf{V}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{m}\mathbf{e}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{s}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}\mathbf{}\left(L\right)}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{3}\mathit{M} \\ \mathit{M}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathit{o}\mathit{f}\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{e}\mathbf{=}\mathit{M}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathit{i}\mathit{t}\mathit{y}\mathbf{\times }\mathit{V}\mathit{o}\mathit{l}\mathit{u}\mathit{m}\mathit{e}\left(L\right)\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{\times }\mathbf{52}\mathbf{=}\mathbf{119}\mathbf{.}\mathbf{6}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathit{e} \end{gathered}$$

Now, $CaC{l}_2$dissociates as

$\mathit{C}\mathit{a}\mathit{C}{\mathit{l}}_{\mathbf{2}}\left(aq\right)\mathbf{\to }\mathit{C}{\mathit{a}}^{\mathbf{2}\mathbf{+}}\left(aq\right)\mathbf{+}\mathbf{2}\mathit{C}\mathit{l}\left(aq\right)$

As 1 mole of $CaC{l}_2$give 1 mole of $C{a}^{2+}$ion so,

Moles of $CaC{l}_2$= Moles of$C{a}^{2+}$ions

1 mole always contains $6.023\times {10}^{23}$particles.

$\mathbf{119}\mathbf{.}\mathbf{6}\mathbf{}\mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{}\mathit{C}{\mathit{a}}^{\mathbf{2}\mathbf{+}}\mathit{i}\mathit{o}\mathit{n}\mathit{s}\mathbf{=}\mathbf{119}\mathbf{.}\mathbf{6}\mathbf{\times }\mathbf{6}\mathbf{.}\mathbf{023}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{25}}$

Thus, number of Cu2+ ions in 52 L of 2.3 M copper(II) chloride isrole="math" localid="1656590428859" $\mathbf{7}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{25}}\mathbf{.}$

Calculation of Molarity of glucose solution

Molarity calculation is easy as volume and number of moles are given. We can use the formula,

$$\begin{gathered} \mathit{M}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathit{i}\mathit{t}\mathit{y}\mathbf{=}\frac{\mathbf{M}\mathbf{o}\mathbf{l}\mathbf{e}\mathbf{s}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{s}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{t}\mathbf{e}}{\mathbf{V}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{m}\mathbf{e}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{s}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}\left(mL\right)}\mathbf{\times }\mathbf{1000} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{135}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}}{\mathbf{275}}\mathbf{\times }\mathbf{1000}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{49}\mathbf{}\mathit{M} \end{gathered}$$

Hence, molarity of 275 mL of solution containing 135mol of glucose is 0.49 M.