The problem asks to calculate how many mL of a 2.6% (m/v) NaNO₃ solution are required to contain 7.4 g of sodium nitrate (NaNO₃). Here's how we can solve this:

Step 1: Understand the concentration formula

The percentage given is a mass/volume (m/v) percentage. This means: $2.6\% \ (\text{m/v}) = \frac{2.6 \ \text{g of NaNO}_3}{100 \ \text{mL of solution}}$

Step 2: Set up the proportion

We are asked how many mL of solution contain 7.4 g of NaNO₃. Using the ratio from the concentration definition, we can set up a proportion: $\frac{2.6 \ \text{g}}{100 \ \text{mL}} = \frac{7.4 \ \text{g}}{x \ \text{mL}}$

Step 3: Solve for $x$

Cross-multiply to solve for $x$: $2.6x = 7.4 \times 100$ $2.6x = 740$ $x = \frac{740}{2.6} = 284.62 \ \text{mL}$

Step 4: Significant Figures

The given data has two significant figures (7.4 g and 2.6%), so the final answer should be rounded to two significant figures: $x \approx 2.8 \times 10^2 \ \text{mL} \ (\text{or } 280 \ \text{mL})$

Final Answer:

The solution would contain approximately 280 mL of the 2.6% NaNO₃ solution to have 7.4 g of sodium nitrate.

Would you like further details on any step? Here are some related questions to consider:

1. How would the calculation change if the concentration were given in molarity instead of percentage?
2. What is the molar mass of sodium nitrate (NaNO₃)?
3. How do you handle more complex percentage concentrations like mass/volume/volume?
4. How can we calculate the mass of solute in a given volume for any percentage concentration?
5. What units would you use if this were a molarity problem instead of mass/volume?

Tip: When working with percentages, always ensure that your units match across the problem (grams with grams, mL with mL).