Answer:
The correct answer is 0.033 M
Explanation:
We have a solution of NaClO with a concentration of 5%w/w:
5% w/w= 5 g NaClO/100 g solution
The first dilution is 10 ml of solution in 100 ml. That is a 1/10 dilution (10ml/100 ml= 1/10). That means we are diluting 10 times the solution. We can calculate the resulting concentration after this first dilution as follows:
5%w/w x 10 ml/100 ml = 5% w/w/10= 0.5%w/w
Then, we take 6 ml of 0.5% w/w solution and we add 6 ml of dye in a reaction vessel. The total volume of the solution in the reaction vessel is 6 ml + 6 ml= 12 ml, and we are diluting twice the solution because 6 ml/12 ml= 1/2. We can calculate the resulting concentration of the solution after this second dilution as follows:
0.5% w/w x 6 ml/12 ml= 0.5% w/w/2= 0.25%w/w
Finally, we need to convert the concentration from %w/w to M (mol solution/1L solution). For this, we assume a density of the solution close to the density of water (1.00 g/ml) and we use the molecular weight of NaClO (74.44 g/mol):
0.25 g NaClO/100 g solution x 1 mol NaClO/74.44 g x 1.00 g solution/1 ml x 100 ml/0.1 L= 0.033 mol/L
= 0.033 M
