Question

What is the concentration of no3- ions in a solution prepared by dissolving 25.0 g of ca(no3)2 in enough water to produce 300. ml of solution?

Answer 1

Ca(NO3)2 --> Ca(+2) + 2 NO3(-1)

Therefore for each mol of Ca(NO3)2 that is dissolved 2 mol of NO3(-1) is put into solution. Now calculate the number of moles of Ca(NO3)2 there is at the start. The molecular weight of Ca(NO3)2 is 164 g/mol, so:

Mol of Ca(NO3)2 = 15.0g/164 g/mol = 0.091 mol of Ca(NO3)2

From the ionic equation, there is 2*0.091 mol of NO3(-1) made in solution, therefore mol of NO3(-1) = 0.182 mol

Now to calculate the concentration in mol/L:

mol of NO3(-1)/volume of sample = 0.182 mol/0.300 L = 0.61 molar

Answer 2

1.016 M is the concentration of nitrate ions in a solution.

Step-by-step explanation:

Molarity of the solution is the moles of compound in 1 Liter solutions.

`Molarity=\frac{\text{Mass of compound}}{\text{Molar mas of compound}* Volume (L)}`

Mass of calcium nitarte = 25.0 g

Molar mass of calcium nitrate = 164 g/mol

Volume of the solution = 300 mL = 0.3 L ( 1 mL =0.001 L)

`Molarity=(25.0 g)/(164 g/mol* 0.3 L)=0.5081 mol/L`

`[Ca(NO_3)_2]=0.5081 M`

`Ca(NO_3)_2(aq)\rightarrow Ca^(2+)(aq)+2NO_3^(-)(aq)`

According to reaction 1 mole of calcium nitrate gives 1 mole of calcium ion and 2 moles of nitrate ions.

`[NO_3^(-)]=2* [Ca(NO_3)_2]=2* 0.5081 M=1.016 M`

1.016 M is the concentration of nitrate ions in a solution.