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Vikki [24]
3 years ago
7

What is the concentration of NO3- ions in a solution prepared by dissolving 20.0 g of Ca(NO3)2 in enough water to produce 300. m

L of solution
Chemistry
1 answer:
Anvisha [2.4K]3 years ago
6 0

Answer:

0.812 M

Explanation:

The reaction that takes place is:

  • Ca(NO₃)₂ → Ca⁺² + 2NO₃⁻

First we <u>convert grams of Ca(NO₃)₂ to moles</u>, using its <em>molecular weight</em>:

  • 20.0 g Ca(NO₃)₂ ÷ 164.088 g/mol = 0.122 mol Ca(NO₃)₂

Then we <u>convert Ca(NO₃)₂ moles to NO₃⁻</u>, keeping in mind the <em>stoichiometry of the reaction</em>:

  • 0.122 moles Ca(NO₃)₂ * \frac{2molNO_{3}^-}{1molCa(NO_3)_2} = 0.244 mol NO₃⁻

Finally we <u>divide the number of moles by the volume</u> (in L), to calculate the concentration:

(300.0 mL ⇒ 300.0 / 1000 = 0.300 L)

  • Molarity = 0.244 mol NO₃⁻ / 0.300 L = 0.812 M
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