Q. 9.117 - An Introduction to General, Organic, and Biological Chemistry

Question

A solution is prepared with $70.0 g of HN{O}_3$ and $130.0 g of H_{2}O$. The $HN{O}_3$ solution has a density of $1.21 g/mL$.

a. What is the mass percent $(m/m)$ of the $HN{O}_3$ solution?

b. What is the total volume, in milliliters, of the solution?

c. What is the mass/volume percent $(m/v)$ of the solution?

d. What is the molarity $\left(M\right)$ of the solution?

Step-by-Step Solution

Answer

(Part a) The Mass percent $(m/m)$ of the $HN{O}_3$ solution is $35\%.$

(Part b) The Total Volume in milliliters of the solution is $163.5 mL.$

(Part c) The mass/volume percent $(m/v)$ of the solution is $42.34\%.$

(Part d) The Molarity of the solution is $6.72 M HN{O}_3.$

Step1: Introduction(Part a)

A mass percentage is a density of solute affirmation that can be approximated from the volume of the solution. The following formula expresses it:

$\text{ Mass percentage of solute }=\frac{\text{ grams of solute }}{\text{ grams of solution }}\times 100\mathrm{\%}$

Step2: Given Information (part a).

Mass of Solute $\left(HN{O}_3\right)=70.0 g$

Mass of Solvent $\left(H_{2}O\right)=130.0 g$

Mass of Solution $(HN{O}_3+H_{2}O)=130.0 g+70.0 g=200 g$

Step3: Explanation (part a).

Determine the Mass percentage:

$\text{ Mass percentage of solute }=\frac{\text{ grams of solute }}{\text{ grams of solution }}\times 100\mathrm{\%}$

$$\begin{gathered} \text{ Mass percent }=\left(\frac{70.0\mathrm{g}}{200.0\mathrm{g}}\right)\times 100\mathrm{\%} \\ =35\% \end{gathered}$$

Step4: Given Information (part b).

Mass of Solute $\left(HN{O}_3\right)=70.0 g$

Mass of Solvent $\left(H_{2}O\right)=130.0 g$

Mass of Solution $(HN{O}_3+H_{2}O)=130.0 g+70.0 g=200 g$

Step5: Explanation (part b).

Calculate the Total Volume in milliliters:

$\text{ density }=\frac{\text{ mass }\left(\mathrm{g}\right)}{\text{ volume }\left(\mathrm{mL}\right)}$

$\mathrm{volume} \left(\mathrm{mL}\right)=\frac{\text{ mass }\left(\mathrm{g}\right)}{\mathrm{density} }$

$$\begin{gathered} \text{ volume }\left(\mathrm{mL}\right)=\frac{200.0\mathrm{g}}{1.21\mathrm{g} {\mathrm{mL}}^{-1}} \\ =165.3 mL \end{gathered}$$

Step6: Given Information (part c).

Mass of Solute $\left(HN{O}_3\right)=70.0 g$

Volume of solution is $163.5 mL.$

Step7: Explanation (part c).

Find the mass/volume percent:

$\text{ Mass percent in }\mathrm{m}/\mathrm{V}=\left(\frac{\text{ mass of solute }}{\text{ Volume of solution }}\right)\times 100\mathrm{\%}$

$$\begin{gathered} \text{ Mass percent }=\left(\frac{70.0\mathrm{g}}{165.3\mathrm{mL}}\right)\times 100\mathrm{\%} \\ \mathrm{Mass}percent= 42.34\% \end{gathered}$$

Step8: Given Information (part d).

Volume of Solution is $163.5 mL$

convert it into Liters,

$$\begin{gathered} 1 L=1000 mL \\ =165.3 m\mathrm{L}\times \frac{1.00\mathrm{L}}{1000 m\mathrm{L}} \\ =0.1653\mathrm{mL} \end{gathered}$$

Step9: Explanation (part d).

Find the Concentration of $HN{O}_3$:

$$\begin{gathered} \text{ Concentration of }{\mathrm{HNO}}_3=\frac{\text{ number of }\mathrm{mol}{\mathrm{HNO}}_3}{\text{ volume in }\mathrm{L}} \\ =\frac{1.11\mathrm{mol} {\mathrm{HNO}}_3}{0.1653 \mathrm{L}} \\ =6.72 M HN{O}_3 \end{gathered}$$