Question

40.0 mL of 2.0 M Fe(NO3)3 is mixed with 2 mL of 5 M Fe(NO3) and 48 mL of water. What is the final molar concentration of Fe(NO3)?

Answer is 1M but how do we get this solution? Please help.

Answer 1

It's simple. We need to find out the total moles of Fe(NO3)3 present in the solution that is newly made.

In the first fraction 40.0 mL of 2.0 M Fe(NO3)3:

2.0 M means 2 moles per 1000ml solution.

Then moles of fe(NO3)3 in 40 ml would be = (40/1000) x 2

= 0.08 moles

In the second fraction, 2 mL of 5 M Fe(NO3)3 :

5 M means 5 moles of fe(NO3)3 in 1000ml of solution,

then moles per 2 ml of soulution would be = (2/1000) x 5

= 0.01

Now, total number of moles of fe(NO3)3 = 0.08 + 0.01 = 0.09 moles

Total volume of the solution that is newly made = 40ml (from fraction 1) + 2ml (from the fraction 2) + 48ml (water)

= 90 ml

total 0.09 moles are present in 90 ml solution.

Molarity = (moles ) x (1000 / volume of the solution in ml)

= 0.09 x (1000 / 90)

= 1M