Q. 9.46 - An Introduction to General, Organic, and Biological Chemistry

Question

Calculate the molarity of each of the following:

a. $0.500$ mole of glucose in a $0.200 L$ of a glucose solution.

b. $73.0 g of HCl$ in a $2.00 L of HCl$ solution.

c. $30.0 g of NaOH$ in $350 mL of NaOH$ solution.

Step-by-Step Solution

Answer

(part a) Molarity of $0.500$mole of glucose in a $0.200 L$ of glucose solution is $2.5 M.$

(part b) Molarity of $73.0 g of HCl$ in $2.00 L of HCl$ solution $1.0 M.$

(part c) Molarity of $30.0 g of NaOH$ in $350 mL of NaOH$ solution is $2.14 M.$

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

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

Step2: Given Information (part a).

Given:

$0.500$ mole of glucose in a $0.200 L$of a glucose solution.

Step3: Explanation (part a).

Substitute the values in formula,

$$\begin{gathered} \mathrm{Molarity} =\frac{\text{ moles of solute }}{\text{ liters of solution }} \\ =\frac{0.500\mathrm{mol}\text{ glucose }}{0.200\mathrm{L}\text{ water }} \\ =2.5 mol/ L \\ =2.5 M \end{gathered}$$

Step4: Given Information (part b).

Given:

$73.0 g of HCl$ in $2.00 L of HCl$ solution.

Step5: Explanation (part b).

Find the moles for $73.0 g of HCl.$

$$\begin{gathered} \text{ number of moles of }\mathrm{HCl}=\text{ mass of }\mathrm{HCl}\times \frac{1\mathrm{mol} \mathrm{HCl}}{\text{ molar mass of }\mathrm{HCl}} \\ =73.0\mathrm{g}\times \frac{1\mathrm{mol} \mathrm{HCl}}{36.46 \mathrm{g}} \\ =2.00 mol HCl \end{gathered}$$

Substitute the values in molarity formula,

$$\begin{gathered} \text{ Molarity }\left(\mathrm{M}\right)\text{ of }\mathrm{HCl}=\frac{2.00\mathrm{mol} \mathrm{HCl}}{2.00\mathrm{L}\text{ water }} \\ =1.0 \mathrm{mol}/\mathrm{L} \\ =1.0 \mathrm{M} \mathrm{HCl} \end{gathered}$$

Step6: Given Information (part c).

Given:

$30.0 g of NaOH$ in $350 mL of NaOH$ solution.

Step7: Explanation (part c).

Find the moles for $30.0 g of NaOH,$

$$\begin{gathered} \text{ number of moles of NaOH}=\text{ mass of }\mathrm{NaOH}\times \frac{1\mathrm{mol} \mathrm{NaOH}}{\text{ molar mass of }\mathrm{NaOH}} \\ =30.0\mathrm{g}\times \frac{1\mathrm{mol} \mathrm{HCl}}{39.997 \mathrm{g}} \\ =0.750 mol \mathrm{NaOH} \end{gathered}$$

Covert milliliter to liter,

$$\begin{gathered} 1000 mL=1 L \\ 350 mL\times \frac{1.00\mathrm{L}}{1000 \mathrm{mL}}=0.350\mathrm{L} \end{gathered}$$

Substitute the values in molarity formula,

$$\begin{gathered} \text{ Molarity }\left(\mathrm{M}\right)\text{ of NaO}H=\frac{2.00\mathrm{mol} \mathrm{NaO}H}{2.00\mathrm{L}\text{ water }} \\ =2.14 \mathrm{mol}/\mathrm{L} \\ =2.14 \mathrm{M} \mathrm{HCl} \end{gathered}$$