To find the mass of the solute present in 685 mL of 0.880 M KBr, we need to follow these steps:
1. Convert the volume from milliliters to liters:
- Given volume = 685 mL
- To convert milliliters to liters, we divide by 1000:
[tex]\[ \text{Volume in liters} = \frac{685 \text{ mL}}{1000} = 0.685 \text{ L} \][/tex]
2. Calculate the number of moles of KBr:
- Given molarity (M) = 0.880 M
- Molarity (M) is defined as the number of moles of solute per liter of solution. We use the formula:
[tex]\[ \text{Moles of KBr} = \text{Molarity} \times \text{Volume in liters} \][/tex]
Substituting the given values:
[tex]\[ \text{Moles of KBr} = 0.880 \text{ M} \times 0.685 \text{ L} = 0.6028 \text{ moles} \][/tex]
3. Calculate the mass of KBr in grams:
- Molar mass of KBr = 119 g/mol
- To find the mass, we multiply the number of moles by the molar mass:
[tex]\[ \text{Mass of KBr} = \text{Moles of KBr} \times \text{Molar mass} \][/tex]
Substituting the calculated moles and given molar mass:
[tex]\[ \text{Mass of KBr} = 0.6028 \text{ moles} \times 119 \text{ g/mol} = 71.7332 \text{ grams} \][/tex]
Therefore, the mass of the solute (KBr) present in 685 mL of 0.880 M KBr is 71.7332 grams.
