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.
Answers
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:
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