A student, who has not learned how to properly prepare solutions, added X g of s
ID: 817346 • Letter: A
Question
A student, who has not learned how to properly prepare solutions, added X g of sucrose(FW = 342.3) to exactly 150.0 mL of dH2O and assumed that a 0.5 M solution had been prepared. The total solution volume increased to 165mL after the the solution was thoroughly mixed and the solid material was dissolved.
(a.) Determine the actual concentration of the solution prepared by the student.
(b.) Determine the percentage error in concentration between the solution actually prepared and the desired 0.5 M concentration.
(c.) If the solution had been prepared correctly to the desired 0.5 M concentration, what volume of this stock should be diluted to prepare 100.0 mL of a 1.0 mM working solution?
Explanation / Answer
*** a ***
the mass of sucrose added would be
150.0mL H2O x (1L / 1000mL) x (0.5 mol sucrose / mL H2O) x (342.3g sucrose / mole sucrose) = 25.672g sucrose
where of course the mistake is
.. (0.5 mol sucrose / L H2O)
it should be
.. (0.5 mol sucrose / L solution)
then.. actual molarity would be
(25.672g sucrose / 165 mL solution) x (1 mol sucrose / 342.3g sucrose) x (1000mL ./ L) = 0.454M
*** b ***
% error = |measured - accepted| / |accepted| x 100% = |0.5 - 0.454| / |0.454| x 100% = 10%
*** c ***
from a mole balance
M1V1 = M2V2
rearranging
.. V1 = V2 x (M2 / M1) = 100.0mL x (1.0mM x 1M / 1000mM / 0.5M) = 0.2mL
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