An experiment requires 15.0 ml of a 0.100 m solution of hcl. if you have a 12.0 m hcl stock solution in the laboratory, how would you prepare enough 0.1 m solution to run at least 3 experiments?

The given choices are as follows
A. add 0.833 mL 12 M HCl stock solution to 99.167 mL water
B.add 0.833 mL 12 M HCl stock solution to 49.167 mL water
C.add 0.833 mL 12 M HCl stock solution to 199.167 mL water

the experiment requires 15.0 mL for each experiment. To conduct at least three experiments the volume needed would be 15.0 x 3 = 45.0 mL
We will need an excess amount of HCl in addition to the 45.0 mL as solutions are required in excess to wash the glassware e.g.: burettes with the solution.
So lets take the required volume of HCl to be prepared as 100.0 mL

now when preparing diluted solutions from stock solutions we can use the dilution equation 
c1v1 = c2v2 
where  c1 = concentration and v1 is volume required from the stock solution 
c2 is concentration and v2 is volume of diluted solution to be prepared
substituting the values in the equation 
12.0 M x V1 = 0.100 M x 100.0 mL 
V1 = 0.833 mL
therefore 0.833 mL should be taken from the 12.0 M stock solution and final volume of 0.100 M HCl to be prepared is 100.0 mL . to make it up to the 100.0 mL mark we need to add (100.0 mL - 0.833 mL ) = 99.167 mL of water.

correct answer is A. add 0.833 mL 12 M HCl stock solution to 99.167 mL water

Number of molecules=(12.01+2x15.9994)/112 x na