Question

25 cm3 of 0.2 M solution of metal choride (MClx) reacted with 150 cm3 of 0.1 M AgNO3 completely to form precipitate of AgCl. What is the formula of metal chloride.

Answer 1

Answer:

To find the number of milli-equivalents, we need to equate.

From the question we get that 25cm3 of 0.2 M Metal chloride will be =150cm3 of 0.1 M of the silver nitrate solution.

Now, 1cm3 of 0.2M of the chloride x will be =  6 cm3 of 0.1 M AgNO3.

​ Then, we get that 1cm3 of 0.1 M of the chloride x will be  = 6 x 0.1/0.2.

                 

Therefore, on solving we get that the value of x =3.

Therefore we can find the formula of the metal chloride which will be MCl3.

Answer 2

Answer:

MCl3

Explanation:

Normality of solution 1 = Normality of solution 2

(N1 * V1 = N2 * V2) ...1

But, Normality = Molarity*n(n-factor) [N1=M1*n] ...2

(I recommend you to know how to find n-factor of salts)

From 1 and 2...

M1 * V1 * n1 = M2 * V2 * n2

Here, M1 = 0.2, V1 = 0.025 L, M2 = 0.1, V2 = 0.150 L

n1 = ? n2 = Ag+ and No3- so n-factor is 1 here

0.2 * 0.025 * n1 = 0.1 * 0.150 * 1

n1 = 3(don't believe me? then try it out)

Hence the N factor is 3

But Metal chloride is M1Clx

where the charges for both are +3 and -3 respectively

Hence here we get to know that the charge of Metal is 3+ because the coefficient is 1.

The charge of Cl is -1(you should know this) and the coefficient(x) is 3 so you can get a -3 charge to balance out the Metals charge.

So the Formula is MCl3