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

Question

You are supplied with 500 mL each of 2N HCl and 5N HCl. What is the maximum volume of 3M HCl that you can prepare using only these two solutions? (A) 250 mL (B) 500 mL (C) 750 mL (D) 1000 mL

Tardigrade Admin · Accepted Answer

As per given relation, N1 V1+N2 V2=N(V1+V2) N1=2 N2=5 V1=500 V2= say -x Thus, 2 × 500+5 x=3(x+500) or 2 x=500 i.e. x=250 Hence, maximum volume of 3 M HCl =500+250=750 mL

Q. You are supplied with $500 m L$ each of $2 N H Cl$ and $5 N H Cl$ . What is the maximum volume of $3 M H Cl$ that you can prepare using only these two solutions?

250 mL

500 mL

750 mL

1000 mL

As per given relation,
$N_{1} V_{1} + N_{2} V_{2} = N (V_{1} + V_{2})$
$N_{1} = 2 N_{2} = 5$
$V_{1} = 500 V_{2} =$ say $- x$
Thus,
$2 \times 500 + 5 x = 3 (x + 500)$ or $2 x = 500$
i.e. $x = 250$
Hence, maximum volume of
$3 M H Cl = 500 + 250 = 750 m L$

Q. You are supplied with 500mL each of 2NHCl and 5NHCl. What is the maximum volume of 3MHCl that you can prepare using only these two solutions?