Question

Question 1 how much water would you add to 175 ml of 6.00 m hydrochloric acid to prepare a 2.00 m solution of this acid? 350 ml 700. ml 58.3 ml 525 ml none of the above

Answer 1

Answer:

350 mL

Explanation:

Given:

concentration of stock solution of HCl =6.00 M

Volume of stock solution of HCl = 175 mL

concentration of required solution = 2.00 M

total volume of required solution = to be determined

This is problem of dilution. We have to determine the total volume of required solution to be made, from the given stock solution.

We will use

M₁V₁=M₂V₂

Where

M₁=concentration of stock solution

V₁= volume of stock solution

M₂=concentration of required solution

V₂=volume of required solution

Putting values

6X175=2XV₂

V₂=525 mL

so total volume of required solution formed will be 525 mL

We have already 175 mL of solution in it. The water need is difference of these two volume.

water added = 525-175= 350 mL

Answer 2

We can use the following dilution equation 
c1v1 = c2v2
where c1 is concentration and v1 is the volume of the concentrated solution 
c2 is concentration and v2 is final volume of the diluted solution 
substituting the values 
6.00 M x 175 mL = 2.00 M x V 
V = 525 mL
final volume of the diluted solution is 525 mL
since there's already 175 mL of acid ,
volume of water to be added - 525 - 175 = 350 mL 
350 mL of water should be added