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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.

Respuesta :

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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