To calculate the molarity $M$ of $\text{NH}_4\text{NO}_3$, we use the formula:

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

Step 1: Calculate the moles of $\text{NH}_4\text{NO}_3$

We are given that 3.72 g of $\text{NH}_4\text{NO}_3$ is dissolved. To convert grams to moles, we use the molar mass of $\text{NH}_4\text{NO}_3$. The atomic masses are approximately:

• $\text{N}$: 14.01 g/mol
• $\text{H}$: 1.01 g/mol
• $\text{O}$: 16.00 g/mol

Thus, the molar mass of $\text{NH}_4\text{NO}_3$ is:

$\text{Molar mass of } \text{NH}_4\text{NO}_3 = (1 \times 14.01) + (4 \times 1.01) + (1 \times 14.01) + (3 \times 16.00) = 80.05 \, \text{g/mol}$

Now, we can calculate the moles of $\text{NH}_4\text{NO}_3$:

$\text{Moles of } \text{NH}_4\text{NO}_3 = \frac{3.72 \, \text{g}}{80.05 \, \text{g/mol}} \approx 0.0465 \, \text{mol}$

Step 2: Convert the volume from mL to L

The volume is given as $5.82 \times 10^2 \, \text{mL}$. To convert milliliters to liters, we divide by 1000:

$\text{Volume in L} = \frac{5.82 \times 10^2 \, \text{mL}}{1000} = 0.582 \, \text{L}$

Step 3: Calculate the molarity

Now, we can use the molarity formula:

$M = \frac{0.0465 \, \text{mol}}{0.582 \, \text{L}} \approx 0.0799 \, \text{mol/L}$

Therefore, the molarity $M$ of the solution is approximately 0.0799 M.

Would you like further details or have any questions about this process?

Related questions:

1. How would you calculate the molarity if the mass of $\text{NH}_4\text{NO}_3$ was different?
2. How does the volume of the solution affect molarity?
3. What happens to molarity if more solute is added without changing the solution volume?
4. How would you approach calculating molarity for a different compound?
5. How can we convert molarity to molality if we know the solution's density?

Tip:

Always ensure the units for volume are in liters when calculating molarity, as molarity is defined in terms of moles per liter.