Respuesta :
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:
1.016 M is the concentration of nitrate ions in a solution.
Explanation:
Molarity of the solution is the moles of compound in 1 Liter solutions.
[tex]Molarity=\frac{\text{Mass of compound}}{\text{Molar mas of compound}\times Volume (L)}[/tex]
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)
[tex]Molarity=\frac{25.0 g}{164 g/mol\times 0.3 L}=0.5081 mol/L[/tex]
[tex][Ca(NO_3)_2]=0.5081 M[/tex]
[tex]Ca(NO_3)_2(aq)\rightarrow Ca^{2+}(aq)+2NO_3^{-}(aq)[/tex]
According to reaction 1 mole of calcium nitrate gives 1 mole of calcium ion and 2 moles of nitrate ions.
[tex][NO_3^{-}]=2\times [Ca(NO_3)_2]=2\times 0.5081 M=1.016 M[/tex]
1.016 M is the concentration of nitrate ions in a solution.
