You are supplied with 500 mL each of 2(N) HCl and 5(N) HCl. What is the maximum volume of 3(M) HCl that you can prepare using only these two solutions-

Options

A. 250 mL
B. 500 mL
C. 750 mL
D. 1000 mL

Answer

C

Explanation

For HCl, normality and molarity are same.
According to the formula, `V_(1)N_(1)+V_(2)N_(2)=V_(3)N_(3)`
`500xx2+x xx5=(500+x)xx3` [Let, x mL of 5(N) HCl is mixed]
`therefore 2x=500 or, x=250 mL`
`therefore` Total volume of the 3(N) HCl solution
=(500+250)mL=750 mL.