Question

A student dissolved 3.00 g of Co(NO 3) 2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the solution then diluted it with water to give 275 mL of a final solution. How many grams of NO 3 - ion are there in the final solution?

Answer 1

0.0812 grams of nitrate ions are there in the final solution.

Step-by-step explanation:

Mass of cobalt (II) nitrate = 3.00 g

Moles of cobalt(II) nitrate =
`(3.00 g)/(183 g/mol)=0.0164 mol`

Volume of the solution = 100 mL = 0.100 L

1 mL = 0.001 L

`Molarity=(Moles)/(Volume(L))`

Molarity of the solution =
`(0.0164 mol)/(0.100 L)=0.164 M`

Cobalt (II) nitrate in its aqueous solution gives 1 mole of cobalt(II) ion and 2 moles of nitrate ions.

`[NO_3^(-)]=2* [Co(NO_3)_2]=2* 0.164 M=0.328 M`

Molarity of the nitrate ion before solution =
`M_1=0.328 M`

Volume of the nitrate ion before solution =
`V_1=4.00 mL`

Molarity of the nitrate ion after solution =
`M_2=?`

Volume of the nitrate ion after solution =
`V_2=275 mL`

`M_1V_1=M_2V_2` ( Dilution)

`M_2=(0.328 M* 4.00 mL)/(275 mL)=0.00477 M`

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

`n=0.00477 M* 0.275 L=0.00131 mol`

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.