Question

4.
Molarity of a solution obtained by mixing 500ml of
0.5M HCl with 200ml of 1M HCI will be
(1) 0.64M
(2) 0.54M
(3) 0.50M
(4) 0.75M​

Answer 1

Answer:

1.will be the correct answer

Answer 2

Answer:

On mixing of 500ml of 0.5M HCl with 200ml of 1M HCI, the molarity of resultant solution is equal to 0.64M.

Therefore, option (1) is correct.

Explanation:

Given, The molarity of first solution, $M_{1}=0.5M$

Volume of the first solution , $V_{1}=500ml=0.5L$

The molarity of second solution, $M_{2}=1M$

Volume of the second solution , $V_{2}=200ml=0.2L$

Molarity of the solution = (no. of moles)/(Volume in L)

$M=\frac{n}{V}$

$n= MV$

Consider that after mixing of 1st and 2nd solution,

Number of moles, $n=n_{1}+n_{2}$                                       ..............(1)

Molarity of resultant solution will be M and volume V= V₁ + V₂

Therefore eq. (1) becomes:-

$M(V_{1}+V_{2})=M_{1}V_{1}+M_{2}V_{2}$

$M(0.5+0.2)=(0.5)(0.5)+(1)(0.2)$

$M(0.7)=0.25+0.2$

$M=\frac{0.45}{0.7}$

$M=0.64$

Therefore, the molarity of resultant solution is 0.64M after mixing of first and second solution.