The volume ratio of 6N and 2N HCl required to prepare 100 mL of 5N HCl is

Question

The volume ratio of 6N and 2N HCl required to prepare 100 mL of 5N HCl is (A) 3: 1 (B) 1: 3 (C) 4: 1 (D) 1: 4

Tardigrade Admin · Accepted Answer

Let texta textmL of text6N and ( text100 - texta) textmL of text2N are mixed So 6a+2(.100-a.)=100× 5 texta text= text75 textmL, ( text100 - text1) text= text25 textmL

Q. The volume ratio of 6N and 2N HCl required to prepare 100 mL of 5N HCl is

3 : 1

1 : 3

4 : 1

1 : 4

Let $a mL$ of $6N$ and $(100 - a) mL$ of $2N$ are mixed

So $6 a + 2 (100 - a) = 100 \times 5$

$a = 75 mL, (100 - 1) = 25 mL$

Q. The volume ratio of 6N and 2N HCl required to prepare 100 mL of 5N HCl is

3 : 1

1 : 3

4 : 1

1 : 4

Let amL of 6N and (100−a)mL of 2N are mixed So 6a+2(100−a)=100×5 a=75mL,(100−1)=25mL

Let $a mL$ of $6N$ and $(100 - a) mL$ of $2N$ are mixed

So $6 a + 2 (100 - a) = 100 \times 5$

$a = 75 mL, (100 - 1) = 25 mL$