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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. mL of solution

Respuesta :

maacastrobr

Answer:

0.812 M

Explanation:

The reaction that takes place is:

  • Ca(NO₃)₂ → Ca⁺² + 2NO₃⁻

First we convert grams of Ca(NO₃)₂ to moles, using its molecular weight:

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

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

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

Finally we divide the number of moles by the volume (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