Question

A student dissolved 4.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution and then diluted it with water to give 275. mL of a final solution. How many grams of NO3- ion are there in the final solution?

Answer 1

Answer:

0.08097 grams of nitrate ions are there in the final solution.

Explanation:

Moles of cobalt(II) nitrate ,n= $\frac{4.00 g}{245 g/mol}=0.01633 mol$

Volume of the cobalt(II) nitrate solution, V = 100.0 mL = 0.1 L

$Molarity=\frac{n}{V(L)}$

Let the molarity of the solution be $M_1$

$M_1=\frac{0.01633 mol}{0.1 L}=0.1633 M$

A students then takes 4 .00 mL of $M_1$ solution and dilute it to 275 ml.

$M_1=0.1633 M$

$V_1=4.00 mL$

$M_2=?$ (molarity after dilution)

$V_2=275 mL$ (after dilution)

$M_1V1=M-2V_2$

$M_2=\frac{M_1V_1}{V_2}=\frac{0.1633 M\times 4.00 mL}{275 mL}=0.002375 M$

Molarity of the of solution after dilution is 0.002375 M.

$Co(NO_3)_2(aq)\rightarrow Co^{2+}(aq)+2NO_3^{-}(aq)$

1 mol of cobalt(II) nitrate gives 2 moles of nitrate ions. Then 0.002375 M solution of cobalt (II) nitrate will give:

$[NO_3^{-}]=\frac{2}{1}\times 0.002375 M=0.004750 M$

Moles of nitrate ions = n

Volume of the solution = 275 mL = 0.275 L

Molarity of the nitrate ions = $[NO_3^{-}]=0.004750 M$

$[NO_3^{-}]=\frac{n}{0.275 L}$

n = 0.001306 mol

Mass of 0.001306 moles of nitrate ions:

0.001306 mol × 62 g/mol= 0.08097 g

0.08097  grams of nitrate ions are there in the final solution.