How many mLs of 28% by weight of NH2(17.0g/mol) with a density of 0.890 g/ml are needed to prepare 500mL?
Of 0.250 M concentration. I got 8.53 but my answer is wrong. How do you get the correct answer ?
- ChemTeamLv 76 years agoFavorite Answer
mass of NH3 required:
(0.250 mol/L) (0.5 L) = x / 17.0 g/mol
x = 2.125 g
mass of 28% NH3 that contains 2.125 g:
2.125 is to y as 28 is to 100
y = 7.589286 g
volume of 7.589286 g of 28% solution:
7.589286 g divided by 0.890 g/mL = 8.53 g
Your answer is correct. So, why was it called wrong? One possibility is that 8.5 is the expected answer (you had too many sig figs for the computer). Another possibility is that the answer key in the computer is wrong. Is there a live person who can access the answer in the computer and confirm/disconfirm this?
Best wishes with your continued studies.Source(s): ChemTeam
- hcbiochemLv 76 years ago
The original stock solution of NH3:
28% means that 100 g of the solution will contain 28 g of NH3. Moles NH3 in that volume is: 28g / 17.0 g/mol = 1.65 moles NH3
The volume of 100 g of solution is: 100 g / 0.89 g/mo = 112 mL or 0.112 L
Molarity of stock solution = 1.65 mol 0.112 L = 14.7 M
M1V1 = M2V2
14.7 M ( V1) = 500 mL (0.25M)
V1 = 8.50 mL