How many moles of solute are in the following solutions?
Part A
250mL of 0.26M acetic acid, CH3CO2H
2.90L of 0.31M NaOH
980mL of 3.5M nitric acid, HNO3
molar mass of CH3COOH
=(12+3+12+16+16+1)
=60
mass of acetic acid in 250 mL of 0.26M
=(250/1000) L * 0.26*60
= 3.9 g
What Mathmate gave you is correct for grams acetic acid in the solution. The problems ask for mols and that is obtained by mols = M x L = ? (in this case just omit the 60).
To find the number of moles of solute in a solution, you need to know the concentration of the solute and the volume of the solution. The formula to calculate the number of moles is:
moles = concentration (M) × volume (L)
Let's apply this formula to each of the given solutions.
Part A - 250mL of 0.26M acetic acid, CH3CO2H:
The given solution volume is 250 mL, which needs to be converted to liters by dividing by 1000.
Volume = 250 mL ÷ 1000 mL/L = 0.25 L
Now, we can calculate the number of moles using the formula:
moles = concentration (M) × volume (L)
moles = 0.26 M × 0.25 L = 0.065 moles
Therefore, there are 0.065 moles of acetic acid, CH3CO2H, in the solution.
Part B - 2.90L of 0.31M NaOH:
The given solution volume is already in liters, so no conversion is needed.
Using the formula:
moles = concentration (M) × volume (L)
moles = 0.31 M × 2.90 L = 0.899 moles
Therefore, there are 0.899 moles of NaOH in the solution.
Part C - 980mL of 3.5M nitric acid, HNO3:
First, convert the given solution volume to liters:
Volume = 980 mL ÷ 1000 mL/L = 0.98 L
Using the formula:
moles = concentration (M) × volume (L)
moles = 3.5 M × 0.98 L = 3.43 moles
Therefore, there are 3.43 moles of nitric acid, HNO3, in the solution.