Question

What is the concentration of no3- ions in a solution prepared by dissolving 15.0 g of ca(no3)2 in enough water to produce 300. ml of solution?

Answer 1

Ca(NO3)2 -------> Ca²⁺ +2NO3⁻

M(Ca(NO3)2)= M(Ca) + M(N) + 6M(O)= 40.0 +14.0 +6*16.0 = 150 g/mol

15.0 g Ca(NO3)2 * 1mol/150 g = 0. 100 mol Ca(NO3)2


Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
1 mol                               2 mol
0.100 mol                           0.200 mol

We have 0.2 mol NO3⁻ in 300. mL=0.300 L of solution,
so
0.200 mol NO3⁻ / 0.300 L solution ≈ 0.667 mol NO3⁻ /L solution = 0.667 M

Concentration of NO3⁻ is 0.667 M.

Answer 2

The concentration of NO₃¯ in the solution is 0.61 M

We'll begin by calculating the number of mole in 15 g of Ca(NO₃)₂. This can be obtained as follow:

Mass of Ca(NO₃)₂ = 15 g

Molar mass of Ca(NO₃)₂ = 40 + 2[14 + (16×3)]

= 40 + 2[14 + 48]

= 40 + 2[62]

= 40 + 124

= 164 g/mol

Mole of Ca(NO₃)₂ =?

Mole = mass / molar mass

Mole of Ca(NO₃)₂ = 15 / 164

Mole of Ca(NO₃)₂ = 0.0915 mole

Next, we shall determine the molarity of Ca(NO₃)₂. This can be obtained as follow:

Mole of Ca(NO₃)₂ = 0.0915 mole  

Volume = 300 mL = 300 / 1000 = 0.3 L

Molarity of Ca(NO₃)₂ =?

Molarity = mole / Volume

Molarity of Ca(NO₃)₂ = 0.0915 / 0.3

Molarity of Ca(NO₃)₂ = 0.305 M

Finally, we shall determine the concentration of NO₃¯ in the 0.305 M Ca(NO₃)₂ solution. This can be obtained as follow:

Ca(NO₃)₂(aq) —> Ca²⁺ (aq) + 2NO₃¯ (aq)

From the balanced equation above,

1 mole of Ca(NO₃)₂ produced 2 moles of NO₃¯.

Therefore,

0.305 M Ca(NO₃)₂  will produce = 2 × 0.305 = 0.61 M NO₃¯

Thus, the concentration of NO₃¯ in the solution is 0.61 M

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