Question

1 M HCL and 2 M HCL are mixed in the volume ratio of 4 :1 what is the final morality of HCL solution.

Answer 1

Final Answer : 1.2 M

Steps :
1) Let the Volume of 2 M HCl solution is V litre.
Then, Volume of 1 M HCl is 4V litre.

2) We know,
Here,
No. of moles of 1 M HCl = 1 * 4V = 4V moles.
No. of moles of 2 M HCl = 2 * 1V = 2V moles.

Molarity = No. of Moles of HCl / Volume of solution
=( 4V + 2 V) / (4V + V )
= 6V / 5V
=6/5 = 1.2M

Answer 2

Answer : The final volume of HCl solution is 3 M.

Solution : Given,

Volume ratio is 4 : 1

Let, the volume of 2 M HCl be 'v' and the volume of 1 M HCl be '4v'.

Formula used :  $Molarity=\frac{moles}{\text{ volume of solution}}$

Total volume of solution = v + 4v = 5v

Now we have to calculate the moles for 1 M HCl solution and 2 M HCl solution.

For 1 M HCl :

Moles of 1 M HCl = Molarity × volume of solution = 1 × 5v = 5v moles

For 2 M HCl :

Moles of 2 M HCl = Molarity × volume of solution = 2 × 5v = 10v moles

Total moles of HCl = 5v + 10v = 15v

Now we have to calculate the final molarity of HCl solution.

$Molarity=\frac{\text{ Total moles}}{\text{ Total volume of solution}}=\frac{15v}{5v}=3M$

Therefore, the final volume of HCl solution is 3 M.