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

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 = $\frac{3.00 g}{183 g/mol}=0.0164 mol$

Volume of the solution = 100 mL = 0.100 L

1 mL = 0.001 L

$Molarity=\frac{Moles}{Volume(L)}$

Molarity of the solution = $\frac{0.0164 mol}{0.100 L}=0.164 M$

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

$[NO_3^{-}]=2\times [Co(NO_3)_2]=2\times 0.164 M=0.328 M$

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)

$M_2=\frac{0.328 M\times 4.00 mL}{275 mL}=0.00477 M$

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

$n=0.00477 M\times 0.275 L=0.00131 mol$

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.