Question

A solution is prepared with \(70.0 \mathrm{~g}\) of \(\mathrm{HNO}_{3}\) and \(130.0 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\). It has a density of \(1.21 \mathrm{~g} / \mathrm{mL}\). a. What is the mass percent \((\mathrm{m} / \mathrm{m})\) of the \(\mathrm{HNO}_{3}\) solution? b. What is the total volume of the solution? c. What is the mass/volume percent \((\mathrm{m} / \mathrm{v}) ?\) d. What is its molarity (M)?

Short answer

a. 35%b. 165.29 mLc. 42.34%d. 6.72 M

Step-by-step solution

Calculate the mass percent (\text{m}/\text{m}) of \text{HNO}_3

The mass percent (\text{m}/\text{m}) is calculated by dividing the mass of \text{HNO}_3 by the total mass of the solution and then multiplying by 100. \[ \text{Mass percent} = \frac{70.0 \text{ g}}{70.0 \text{ g} + 130.0 \text{ g}} \times 100 = \frac{70.0}{200.0} \times 100 = 35\text{ \text{ %}} \]

Calculate the total volume of the solution

To find the total volume, use the density formula: \[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \] Rearrange to solve for volume: \[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \] The total mass of the solution is \[ 70.0 \text{ g} + 130.0 \text{ g} = 200.0 \text{ g} \] Therefore, \[ \text{Volume} = \frac{200.0 \text{ g}}{1.21 \text{ g/mL}} = 165.29 \text{ mL} \]

Calculate the mass/volume percent (\text{m}/\text{v})

The mass/volume percent (\text{m}/\text{v}) is found by dividing the mass of the solute by the volume of the solution and then multiplying by 100. \[ \text{Mass/volume percent} = \frac{70.0 \text{ g}}{165.29 \text{ mL}} \times 100 = 42.34\text { %} \]

Calculate the molarity (\text{M}) of the solution

First, convert the mass of \text{HNO}_3 to moles using its molar mass. The molar mass of \text{HNO}_3 is: \[ 1 \text{ H} (1 \text{ g/mol}) + 1 \text{ N} (14 \text{ g/mol}) + 3 \text{ O} (16 \text{ g/mol each}) = 63 \text{ g/mol} \] The number of moles of \text{HNO}_3 is: \[ \frac{70.0 \text{ g}}{63 \text{ g/mol}} = 1.11 \text{ moles} \] Molarity (\text{M}) is calculated by dividing the number of moles by the volume in liters: \[ \text{Volume in liters} = \frac{165.29 \text{ mL}}{1000} = 0.1653 \text { L} \] Therefore, \[ \text{Molarity } (\text {M}) = \frac{1.11 \text{ moles}}{0.1653 \text{ L}} = 6.72 \text{ M} \]

Key concepts

mass percent

Mass percent (commonly denoted as % m/m) indicates the mass of the solute divided by the total mass of the solution, multiplied by 100. This gives us the percentage of the solution's mass that is made up of the solute. Let’s break it down:
Example: Suppose you have a solution with 70.0 g of \text{HNO}_3 (Nitric acid) and 130.0 g of water. The total mass of the solution is the sum of the masses of the solute and solvent.
The formula for mass percent is:
$$\text{Mass percent} = \frac{\text{Mass of solute}}{\text{Total mass of solution}} \times 100$$
Using the provided data:
$$\text{Mass percent} = \frac{70.0 \text{ g}}{70.0 \text{ g} + 130.0 \text{ g}} \times 100 = \frac{70.0}{200.0} \times 100 = 35.0 \text{ \text{ %}}$$
Hence, the mass percent of \text{HNO}_3 in the solution is 35%.
This means 35% of the solution's total mass is \text{HNO}_3.

density and volume

Density relates a substance's mass to its volume and is expressed as mass per unit volume (\text{g/mL}, \text{kg/L}, etc.).
The formula to calculate density is:
$$\text{Density} = \frac{\text{Mass}}{\text{Volume}}$$
To find the total volume of a solution, you can rearrange the formula:
$$\text{Volume} = \frac{\text{Mass}}{\text{Density}}$$
In the exercise, the total mass of the solution is 200.0 g, and its density is 1.21 \text{g/mL}. Therefore, the volume is calculated as:
$$\text{Volume} = \frac{200.0 \text{ g}}{1.21 \text{ g/mL}} = 165.29 \text{ mL}$$
This means the total volume of the \text{HNO}_3 solution is 165.29 mL.

mass/volume percent

Mass/volume percent (\text{m}/\text{v}) is a way of expressing the concentration of a solution. It denotes the mass of the solute in grams per 100 mL of solution.
The formula for mass/volume percent is:
$$\text{Mass/volume percent} = \frac{\text{Mass of solute (g)}}{\text{Volume of solution (mL)}} \times 100$$
Using the previous data:
$$\text{Mass/volume percent} = \frac{70.0 \text{ g}}{165.29 \text{ mL}} \times 100 = 42.34 \text{ \text{ %}}$$
This shows that 42.34% of the solution's volume is \text{HNO}_3.

molarity

Molarity (\text{M}) indicates the number of moles of solute per liter of solution. It is a key concept in chemistry for measuring the concentration of solutions.
To calculate molarity:

• First, determine the number of moles of the solute: $$\text{Moles} = \frac{\text{Mass of solute (g)}}{\text{Molar mass (g/mol)}}$$
  Here, the molar mass of \text{HNO}_3 is 63 \text{ g/mol}. Therefore, the moles of \text{HNO}_3 are:
  $$\text{Moles of } \text{HNO}_3 = \frac{70.0 \text{ g}}{63 \text{ g/mol}} = 1.11$$
• Next, convert the solution volume from mL to L:$$\text{Volume (L)} = \frac{165.29 \text{ mL}}{1000} = 0.1653 \text{ L}$$
• Finally, calculate molarity:$$\text{Molarity} (\text {M}) = \frac{\text{Moles of solute}}{\text{Volume of solution (L)}} = \frac{1.11 \text{ moles}}{0.1653 \text{ L}} = 6.72 \text{ M}$$

This means the molarity of the \text{HNO}_3 solution is 6.72 M. This shows that 6.72 moles of \text{HNO}_3 are present in every liter of the solution.