Question

A stock solution of 0.200 M Fe(NO3)3 solution is prepared in a 100 mL volumetric flask with 1 M HNO3. 10 mL of this stock solution is then pipetted into a beaker and 40 mL of HNO3 is added. This new solution is solution A. Calculate the actual concentration of Fe(NO3)3, determine the number of moles delivered, and the molarity of solution A. 10 mL of solution A is then pipetted into another beaker and 15mL of HNO3 is added. This new solution is solution B. Calculate the same things for solution B.

Answer 1

Solution.

The molar concentration can be calculated by a formula c = n/V. The amount of substance then n = cV.

Preparation of solution A.

10 mL of stock solution contains n = (0.200 M)(0.010 L) = 0.002 moles of Fe(NO₃)₃ delivered.

The actual concentration of Fe(NO₃)₃ in the solution A is (the total volume is used) c = 0.002/(0.01+0.04) = 0.04 M.

Preparation of solution B.

10 mL of solution A contains n = (0.04 M)(0.010 L) = 0.0004 moles of Fe(NO₃)₃ delivered.

The actual concentration of Fe(NO₃)₃ in the solution B is c = 0.0004/(0.01+0.015) = 0.016 M.