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

To determine how many milliliters of a 1.55 M NaBr solution contain 75.0 g of NaBr, we need to follow these steps:

1. Find the Molar Mass of NaBr:

Sodium bromide (NaBr) has a molar mass of 102.89 g/mol. This value represents the mass of one mole of NaBr.

2. Calculate the Number of Moles of NaBr:

To find the number of moles, we use the formula:

[tex]\[
\text{moles of NaBr} = \frac{\text{mass of NaBr (g)}}{\text{molar mass of NaBr (g/mol)}}
\][/tex]

Given that the mass of NaBr is 75.0 g:

[tex]\[
\text{moles of NaBr} = \frac{75.0 \, \text{g}}{102.89 \, \text{g/mol}} \approx 0.729
\][/tex]

3. Determine the Volume of the Solution:

The molarity formula relates the moles of solute, volume of solution in liters, and molarity:

[tex]\[
\text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}
\][/tex]

Plugging the values into the formula with the molarity given as 1.55 M, we find:

[tex]\[
\text{volume of solution (L)} = \frac{\text{moles of NaBr}}{\text{Molarity}} = \frac{0.729 \, \text{moles}}{1.55 \, \text{M}} \approx 0.470 \, \text{L}
\][/tex]

4. Convert Volume from Liters to Milliliters:

Since there are 1000 milliliters in a liter, convert the volume to milliliters:

[tex]\[
\text{volume in milliliters} = 0.470 \, \text{L} \times 1000 \approx 470.28 \, \text{mL}
\][/tex]

Thus, 470.28 mL of a 1.55 M NaBr solution contains 75.0 g of NaBr.