Answer :
To find out how many milliliters of a 1.55 M NaBr solution contain 75.0 g of NaBr, follow these steps:
1. Calculate 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.
- Add these together to get the molar mass of NaBr:
[tex]\(22.99 + 79.90 = 102.89 \, \text{g/mol}\)[/tex].
2. Determine the number of moles of NaBr in 75.0 g:
- Use the formula: [tex]\(\text{moles} = \frac{\text{mass in grams}}{\text{molar mass}}\)[/tex].
- Substitute the values:
[tex]\(\text{moles of NaBr} = \frac{75.0 \, \text{g}}{102.89 \, \text{g/mol}} \approx 0.7289 \, \text{moles}\)[/tex].
3. Calculate the volume of the solution:
- Use the formula: [tex]\(\text{volume (L)} = \frac{\text{moles of solute}}{\text{molarity}}\)[/tex].
- Given that the molarity (concentration) is 1.55 M (mol/L), substitute the values:
[tex]\(\text{volume (L)} = \frac{0.7289 \, \text{moles}}{1.55 \, \text{mol/L}} \approx 0.4703 \, \text{L}\)[/tex].
4. Convert the volume from liters to milliliters:
- Since there are 1000 milliliters in a liter, convert the volume:
[tex]\(0.4703 \, \text{L} \times 1000 \, \text{mL/L} = 470.3 \, \text{mL}\)[/tex].
Therefore, you would need approximately 470.3 milliliters of a 1.55 M NaBr solution to contain 75.0 g of NaBr.
1. Calculate 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.
- Add these together to get the molar mass of NaBr:
[tex]\(22.99 + 79.90 = 102.89 \, \text{g/mol}\)[/tex].
2. Determine the number of moles of NaBr in 75.0 g:
- Use the formula: [tex]\(\text{moles} = \frac{\text{mass in grams}}{\text{molar mass}}\)[/tex].
- Substitute the values:
[tex]\(\text{moles of NaBr} = \frac{75.0 \, \text{g}}{102.89 \, \text{g/mol}} \approx 0.7289 \, \text{moles}\)[/tex].
3. Calculate the volume of the solution:
- Use the formula: [tex]\(\text{volume (L)} = \frac{\text{moles of solute}}{\text{molarity}}\)[/tex].
- Given that the molarity (concentration) is 1.55 M (mol/L), substitute the values:
[tex]\(\text{volume (L)} = \frac{0.7289 \, \text{moles}}{1.55 \, \text{mol/L}} \approx 0.4703 \, \text{L}\)[/tex].
4. Convert the volume from liters to milliliters:
- Since there are 1000 milliliters in a liter, convert the volume:
[tex]\(0.4703 \, \text{L} \times 1000 \, \text{mL/L} = 470.3 \, \text{mL}\)[/tex].
Therefore, you would need approximately 470.3 milliliters of a 1.55 M NaBr solution to contain 75.0 g of NaBr.
