Your 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 solution-

Options

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

Explanation

For HCl solution, Normality (N) = Molarity (M) As 2 (N) < 3 (M) and 3 (M) < 5 (N), volume of 3 (M) solution will be maximum if 2 (N) solution is taken i maximum amount i.e., 500 mL. Let, VmL of 5 (N) solution is taken.
`:.V_1V_1+N_2V_2=N_3V_3`
or `2xx500+5xxV=3xx(500+V)`
or 2V=500 or V=250
Therefore, maximum volume of 3 (M) solution
(500 + 250) = 750 mL.