Q3.96P - Chemistry: Molecular Nature Of Matter And Change

Question

Calculate each of the following quantities:

(a) Grams of solute in 185.8 mL of 0.267 M calcium acetate 

(b) Molarity of 500. mL of solution containing 21.1 g of potassium iodide 

(c) Moles of solute in 145.6 L of 0.850 M sodium cyanide

Step-by-Step Solution

Answer

Answer

  (a) Mass of solute in 185.8 mL of 0.267 M calcium acetate is 7.84 g.

  (b) Molarity of 500. mL of solution containing 21.1 g potassium iodideis 0.254M.

  (c) Moles of solute in 145.6 L of 0.850 M sodium cyanideare 123.76 mole.

Step 1: Mass of solute calculation

Molarity is given by the formula,

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

Using the given data, we can say

$0.267=\frac{Moles of solute}{185.8}\times 1000 Moles of solute=\frac{0.267\times 185.8}{1000}=0.0496 mole$

Mass can be calculated as:

$$\begin{gathered} \mathit{M}\mathit{a}\mathit{s}\mathit{s}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{e}\mathbf{=}\mathit{M}\mathit{o}\mathit{l}\mathit{e}\mathbf{\times }\mathit{M}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathbf{ }\mathit{m}\mathit{a}\mathit{s}\mathit{s} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0496}\mathbf{\times }\mathbf{158}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{84}\mathbf{ }\mathit{g} \end{gathered}$$

Hence, mass of solute is 7.84 g.

Step 2: Calculation of Molarity of Potassium Iodide solution

As we have already checked the formula of molarity calculation so we know that we need moles and volume for molarity calculation. Volume is already given so now moles can be calculated as:

$\mathit{M}\mathit{o}\mathit{l}\mathit{e}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{e}\mathbf{=}\frac{\mathbf{M}\mathbf{a}\mathbf{s}\mathbf{s}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{s}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{t}\mathbf{e}}{\mathbf{M}\mathbf{o}\mathbf{l}\mathbf{a}\mathbf{r}\mathbf{ }\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}}\mathbf{=}\frac{\mathbf{21}\mathbf{.}\mathbf{1}}{\mathbf{166}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{127}$

Now, molarity can be easily calculated

$\mathit{M}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathit{i}\mathit{t}\mathit{y}\mathbf{=}\frac{\mathbf{M}\mathbf{o}\mathbf{l}\mathbf{e}\mathbf{s}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{s}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{t}\mathbf{e}}{\mathbf{V}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{m}\mathbf{e}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{s}\mathbf{o}\mathbf{l}\mathbf{i}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}\left(mL\right)}\mathbf{\times }\mathbf{1000}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{127}}{\mathbf{500}\mathbf{ }\mathbf{m}\mathbf{L}}\mathbf{\times }\mathbf{1000}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{254}\mathbf{ }\mathit{M}$

Thus, molarity of the given solution is 0.254 M.

Step 3: Calculation of Moles of solute insodium cyanide solution

Moles calculation is easy as volume and molarity are given. We can use the formula,

$$\begin{gathered} \mathit{M}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathit{i}\mathit{t}\mathit{y}\mathbf{=}\frac{\mathbf{M}\mathbf{o}\mathbf{l}\mathbf{e}\mathbf{s}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{s}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{t}\mathbf{e}}{\mathbf{V}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{m}\mathbf{e}\mathbf{ }\mathbf{o}\mathbf{f}\mathbf{ }\mathbf{s}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}\left(L\right)} \\ \mathit{M}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{e}\mathbf{=}\mathit{M}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathit{i}\mathit{t}\mathit{y}\mathbf{\times }\mathit{V}\mathit{o}\mathit{l}\mathit{u}\mathit{m}\mathit{e}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\left(L\right) \\ \mathit{M}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{e}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{850}\mathbf{\times }\mathbf{145}\mathbf{.}\mathbf{6}\mathbf{=}\mathbf{123}\mathbf{.}\mathbf{76}\mathbf{ }\mathit{m}\mathit{o}\mathit{l}\mathit{e} \end{gathered}$$

Moles of solute in given solution of sodium cyanide are 123.76 mole.