Q. 9.45 - An Introduction to General, Organic, and Biological Chemistry

Question

Calculate the molarity of each of the following:

a. $2.00$ moles of glucose in $4.00 L$ of a glucose solution.

b. $4.00 g of KOH$ in $2.00 L of a KOH$ solution.

c. $5.85 g of NaCl$ in $400 mL of a NaCl$ solution.

Step-by-Step Solution

Answer

(part a) Molarity of$2.00$ moles of a glucose solution in$4.00 L$ of a glucose solution is $0.500 M.$

(part b) Molarity of $4.00 g of KOH$ in $2.00 L of a KOH$ solution is $0.0357 M.$

(part c) Molarity of $5.85 g of NaCl$ in $400 mL of a NaCl$ solution is $0.250 M.$

Step1: Introduction (part a).

The molarity of a solution is just the molar mass of solute inside one liter of solution. Molarity is mathematically expressed as

$\text{ Molarity }=\frac{\text{ moles of solute }}{\text{ liters of solution }}$

Step2: Find the molarity (part a).

Given:

$2.00$ moles of a glucose solution in $4.00 L$ of a glucose solution

Substitute the values in formula,

$$\begin{gathered} \text{ Molarity }=\frac{\text{ moles of glucose }}{\text{ liters of solution }} \\ =\frac{2.00\text{ moles }}{4.00\mathrm{L}} \\ =0.500 M \end{gathered}$$

Step3: Find the Molarity (part b).

Given:

$4.00 g of KOH$ in $2.00 L of a KOH$ solution.

Find the moles for $4.00 g of KOH,$

$1\text{ mole of }\mathrm{KOH}=56.1\mathrm{g}$

And then,

$\frac{1\text{ mole }\mathrm{KOH}}{56.1 g \mathrm{KOH}}\text{ and }\frac{56.1 g \mathrm{KOH}}{1\text{ mole }\mathrm{KOH}}$

Hence, Convert it to find the moles for grams,

$$\begin{gathered} \text{ Moles of }\mathrm{KOH}=4.00 \mathrm{g} KOH\times \frac{1\text{ moles }\mathrm{KOH}}{56.1 g \mathrm{KOH}} \\ =0.0710 mole \end{gathered}$$

Substitute the values in molarity formula,

$$\begin{gathered} \text{ Molarity }=\frac{\text{ moles of glucose }}{\text{ liters of solution }} \\ =\frac{0.071\text{ moles }}{2.00 \mathrm{L}} \\ =0\text{.0357 M} \end{gathered}$$

Step4: Find the Molarity (part c).

Given:

$5.85 g of NaCl$ in $400 mL of a NaCl$ solution.

Find the moles for $5.85 g of NaCl,$

$1\text{ mole of }\mathrm{NaCl}=58.4 \mathrm{g}$

And then,

$\frac{1\text{ mole }\mathrm{NaCl}}{58.4g \mathrm{NaCl}}\text{ and }\frac{58.4g \mathrm{NaCl}}{1\text{ mole }\mathrm{NaCl}}$

Hence, Convert it to find the moles for grams,

$$\begin{gathered} \text{ Moles of }\mathrm{NaCl}=5.85 g \mathrm{NaCl}\times \frac{1\text{ moles }\mathrm{NaCl}}{58.4 g \mathrm{NaCl}} \\ =0.100 M \end{gathered}$$

Substitute the values in molarity formula,

$$\begin{gathered} \text{ Molarity }=\frac{\text{ moles of }\mathrm{NaCl}}{\text{ liters of solution }} \\ =\frac{0.100\text{ moles }}{400 \mathrm{mL}\times \frac{1\mathrm{L}}{1000\mathrm{mL}}} \\ =0.250 M \end{gathered}$$