Respuesta :
Answer:
0.0812 grams of nitrate ions are there in the final solution.
Explanation:
Mass of cobalt (II) nitrate = 3.00 g
Moles of cobalt(II) nitrate = [tex]\frac{3.00 g}{183 g/mol}=0.0164 mol[/tex]
Volume of the solution = 100 mL = 0.100 L
1 mL = 0.001 L
[tex]Molarity=\frac{Moles}{Volume(L)}[/tex]
Molarity of the solution = [tex]\frac{0.0164 mol}{0.100 L}=0.164 M[/tex]
Cobalt (II) nitrate in its aqueous solution gives 1 mole of cobalt(II) ion and 2 moles of nitrate ions.
[tex][NO_3^{-}]=2\times [Co(NO_3)_2]=2\times 0.164 M=0.328 M[/tex]
Molarity of the nitrate ion before solution = [tex]M_1=0.328 M[/tex]
Volume of the nitrate ion before solution = [tex]V_1=4.00 mL[/tex]
Molarity of the nitrate ion after solution = [tex]M_2=?[/tex]
Volume of the nitrate ion after solution = [tex]V_2=275 mL[/tex]
[tex]M_1V_1=M_2V_2[/tex] ( Dilution)
[tex]M_2=\frac{0.328 M\times 4.00 mL}{275 mL}=0.00477 M[/tex]
Moles of nitrate ions in 275 ml = n
Molarity of the nitrate ion after solution =0.00477 M
volume of the final solution = 275 mL = 0.275 L
[tex]n=0.00477 M\times 0.275 L=0.00131 mol[/tex]
Mass of 0.00131 moles of nitrate ions:
0.00131 mol × 62 g/mol = 0.0812 g
0.0812 grams of nitrate ions are there in the final solution.
