Calculate the concentration of ammonium nitrate in a solution prepared by dissolving 3.20 g of the salt in enough water to make
100. mL of solution, then diluting 2.00 mL of this solution to a volume of 25.00 mL.
1 answer:
Answer:
.032 M .
Explanation:
Molecular weight of ammonium nitrate is 80 .
3.2 g = 3.2 / 80 moles
= .04 moles
volume = 100 mL = 0.1 L
Molarity of 100 mL solution = .04 moles / 0.1 L
= 0.4 M solution.
Now 2 mL solution of 0.4 M is diluted to a volume of 25 mL .
Using the formula S₁ V₁ = S₂V₂
0.4 M x 2 mL = S₂ x 25 mL
S₂ = .4 x 2 / 25
= .032 M
Hence required concentration is .032 M .
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