Question

What volume of 0.8 m alcl3 solution should be mixed with 50 ml of 0.2m cacl2 solution to get solution of chloride ion concentration equal to 0.6 m?

Answer 1

Heya!!

●Required volume = V

Now,

●Number of moles of AlCl3 = Vx0.8

●Number of moles of Cl- in AlCl3 (n1)

:-

3 x V x 0.8

●Number of moles of Cl- in CaCl2 (n2) :-

(2 x 50 x 0.2)/ 1000

Therefore By The Problem:-

0.6 = (n1+n2)/(v + 50/1000)

Now,

= (2.4-1.6)v= 0.02-0.03

=>1.8v = 0.01

=V = 5.55 × 10^-³ L = 5.55 mL

So, the volume is 55.55 mL.

Thanks!

Answer 2

Answer: The volume of $AlCl_3$ solution is 5.55 mL

Explanation:

To calculate the volume of the solution when 2 solutions are mixed, we use the equation:

$M=\frac{n_1M_1V_1+n_2M_2V_2}{V_1+V_2}$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of the chloride ions in $AlCl_3$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of the chloride ions in $CaCl_2$

1 mole of $AlCl_3$ produces 3 moles of chloride ions and 1 mole of $CaCl_2$ produces 2 moles of chloride ions

We are given:

$n_1=3\\M_1=0.8M\\V_1=?mL\\n_2=2\\M_2=0.2\\V_2=50mL\\M=0.6M$  

Putting all the values in above equation, we get:

$0.6=\frac{(3\times 0.8\times V_1)+(2\times 0.2\times 50)}{V_1+50}\\\\V_1=5.55mL$

Hence, the volume of $AlCl_3$ solution is 5.55 mL