A student wishes to prepare 300 mL of 0.25 M sucrose. He puts 25.7 g of sucrose in a beaker and then adds
300 mL of dH2O. The total volume of the solution after the sucrose has been dissolved is 330 mL.
a) What did he do wrong when preparing his solution?
b) What is the actual concentration of the sucrose solution he prepared?
c) What is the percentage error in the concentration of sucrose from the desired 0.25 M?
He should have diluted to 300 mL of _solution_ as opposed to just adding 300 mL of water.
For B, use this:
MV = mass / molar mass
(x) (0.330 L) = 25.7 g / 342.2948 g/mol
x = 0.2275 M
Round to two sig figs is 0.23 M
For C, the difference is 0.02 between the 0.23 and 0.25. The percent error this this:
(0.02 / 0.25) * 100 = 8%
a) The student made a mistake by adding water after adding the sucrose. It should be done the other way around - first dissolved the sucrose in water and then adjust the final volume.
b) To calculate the actual concentration of the sucrose solution, we need to find the amount of sucrose dissolved in the given volume of water. Since 25.7 g of sucrose was added to 300 mL of water, the concentration can be calculated as follows:
Concentration = (Amount of solute / Volume of solution) * (1000 / MW of solute)
= (25.7 g / 300 mL) * (1000 / 342.3 g/mol)
= 0.0898 M
Therefore, the actual concentration of the sucrose solution is 0.0898 M.
c) The percentage error in the concentration of sucrose can be calculated by comparing the actual concentration to the desired concentration:
Percentage error = [(Actual concentration - Desired concentration) / Desired concentration] * 100
= [(0.0898 M - 0.25 M) / 0.25 M] * 100
= -64.08%
Therefore, the percentage error in the concentration of sucrose from the desired 0.25 M is approximately -64.08%.