A 1.250 g mass of standard aspirin is dissolved in approximately 20 mL of 1 M NaOH. This solution is added to a 250-mL volumetric flask and diluted to the mark with DI water. How many grams per mL of aspirin are in the resulting solution? (Note that a 250-mL volumetric flask has a precision of 0.01 mL and the amount of NaOH added has no bearing on the significant figures in the final concentration.) (g per mL)
1.250/250 = ?
.005
To calculate the concentration of aspirin in the resulting solution, we first need to calculate the number of moles of aspirin present.
1.250 g of aspirin is dissolved in approximately 20 mL of 1 M NaOH. We can assume that the aspirin completely reacts with NaOH to form a salt, so the number of moles of aspirin present is equal to the number of moles of NaOH.
First, let's calculate the number of moles of NaOH:
moles of NaOH = concentration (in moles/L) × volume (in Liters)
moles of NaOH = 1 M × 0.020 L
moles of NaOH = 0.020 moles
Since the number of moles of aspirin is the same as the number of moles of NaOH, we have 0.020 moles of aspirin.
Next, we need to calculate the volume of the resulting solution in mL:
The solution is transferred to a 250-mL volumetric flask and diluted to the mark. The volume of the resulting solution is 250 mL.
Finally, we can calculate the concentration of aspirin in the resulting solution in grams per mL:
concentration (in g/mL) = mass (in grams) / volume (in mL)
concentration (in g/mL) = 1.250 g / 250 mL
concentration (in g/mL) = 0.005 g/mL
Therefore, the concentration of aspirin in the resulting solution is 0.005 g/mL.