How many milliliters of a 1.55 M NaBr solution contain 75.0 g of NaBr?

To find out how many milliliters of a 1.55 M NaBr solution contain 75.0 grams of NaBr, follow these steps:

1. Find the Molar Mass of NaBr:
- Sodium (Na) has an atomic mass of approximately 22.99 g/mol.
- Bromine (Br) has an atomic mass of approximately 79.90 g/mol.
- Molar mass of NaBr = 22.99 g/mol + 79.90 g/mol = 102.89 g/mol.

2. Calculate Moles of NaBr:
- To determine the number of moles, use the formula:
[tex]\[
\text{moles of NaBr} = \frac{\text{mass of NaBr}}{\text{molar mass of NaBr}}
\][/tex]
- With 75.0 grams of NaBr, the moles are:
[tex]\[
\text{moles of NaBr} = \frac{75.0 \, \text{g}}{102.89 \, \text{g/mol}} \approx 0.7289 \, \text{moles}
\][/tex]

3. Determine the Volume of Solution in Liters:
- Since the concentration of the solution is given as 1.55 M (moles per liter), use the formula:
[tex]\[
\text{volume in liters} = \frac{\text{moles of NaBr}}{\text{concentration in M}}
\][/tex]
- With 0.7289 moles and a 1.55 M concentration:
[tex]\[
\text{volume in liters} \approx \frac{0.7289 \, \text{moles}}{1.55 \, \text{M}} \approx 0.4703 \, \text{liters}
\][/tex]

4. Convert Volume to Milliliters:
- Since 1 liter equals 1000 milliliters, convert the volume:
[tex]\[
\text{volume in milliliters} = 0.4703 \, \text{liters} \times 1000 \, \text{mL/L} \approx 470.3 \, \text{mL}
\][/tex]

Therefore, approximately 470.3 milliliters of the 1.55 M NaBr solution are needed to contain 75.0 grams of NaBr.