To find the mass of $\text{NH}_3$ required to make a 0.250 molal ($m$) solution in 475 g of methanol, we can use the definition of molality:

$\text{Molality (m)} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$

Step 1: Rearrange the formula to solve for moles of $\text{NH}_3$

$\text{Moles of } \text{NH}_3 = \text{Molality (m)} \times \text{Mass of solvent in kg}$

Step 2: Convert the mass of methanol from grams to kilograms

$\text{Mass of solvent in kg} = \frac{475 \text{ g}}{1000} = 0.475 \text{ kg}$

Step 3: Calculate the moles of $\text{NH}_3$

$\text{Moles of } \text{NH}_3 = 0.250 \, m \times 0.475 \, \text{kg} = 0.11875 \, \text{moles}$

Step 4: Calculate the molar mass of $\text{NH}_3$

• Nitrogen (N): 14.01 g/mol
• Hydrogen (H): 1.008 g/mol (there are 3 hydrogen atoms)

$\text{Molar mass of } \text{NH}_3 = 14.01 + (3 \times 1.008) = 17.034 \text{ g/mol}$

Step 5: Calculate the mass of $\text{NH}_3$

$\text{Mass of } \text{NH}_3 = \text{Moles of } \text{NH}_3 \times \text{Molar mass of } \text{NH}_3$

$\text{Mass of } \text{NH}_3 = 0.11875 \, \text{moles} \times 17.034 \, \text{g/mol} \approx 2.022 \, \text{g}$

So, the mass of $\text{NH}_3$ that must be dissolved in 475 g of methanol to make a 0.250 $m$ solution is approximately 2.022 g.

Would you like more details or have any questions about this calculation?

Here are five related questions you might find helpful:

1. How do you calculate the molality of a solution given the mass of solute and solvent?
2. How does the density of a solvent affect the preparation of solutions?
3. What is the difference between molality and molarity?
4. How would you prepare a solution with a specific molality using a different solute?
5. What is the effect of temperature on the molality of a solution?

Tip: Molality is independent of temperature, making it particularly useful in situations where temperature varies.