Question

How much NaC₂H₃O₂ •3H₂O would have to be weighed out to prepare   100 mL   of a   0.25 M   solution? Report your answer to three significant figures.

Answer: 3.40

If you weighed out 3.50 percent less NaC₂H₃O₂ • 3H₂O than required, what would be the molarity of the solution?

Answer 1

no. of mole = molarity X volume of solution in liter

100 ml = 0.100 liter

no. of mole of NaC₂H₃O₂.3H₂O = 0.25 X 0.100 = 0.025 mole

molar mass of NaC₂H₃O₂.3H₂O = 136 g/mol then 0.025 mole of NaC₂H₃O₂.3H₂O = 136 X 0.025 = 3.40 gm

Answer = 3.40 gm

3.40 gm = 100 % then 3.50 % percent = 3.40 X 3.50 / 100 = 0.119 gm

when we add 0.119 gm less that mean we add 3.40 - 0.119 = 3.281 gm

molar mass of NaC₂H₃O₂.3H₂O = 136 g/mol then 3.281 gm = 3.281 / 136 = 0.0241 mole

molarity = no. of mole / volume of solution in liter

molarity of NaC₂H₃O₂.3H₂O = 0.0241 / 0.100 = 0. 241 M

If you weighed out 3.50 percent less NaC₂H₃O₂.3H₂O than required molarity of the solution = 0.241 M