Question

0.18g of a divalent metal was completely dissolved in 250 cc of acid solution containing 4.9 g H2SO4 per liter. 50 cc of the residual acid solution required 20 cc of N/10 alkali for complete neutralization. Calculate the atomic weight of metal.​

Answer 1

Answer:

the atomic weight of metal is 5.08g.

Answer 2

Given: Weight of divalent metal = 0.18g

Volume of acid solution (V1) = 250cc or 250ml

Volume of acid solution left (V2) = 50cc or 50 ml

Normality of the alkali solution used (N2) = N / 10

The volume of the alkali solution (V3) = 20cc or 20 ml

To Find: atomic weight of metal.​

Solution:

Let the normality of the acid be N1

According to the normality equation, 'The product of normality and volume of the reacting acid and base is equal.

So, N1 V1 = N2 V2, where N1 and V1 are the normality and volume of the acid, and N2, V2 are the normality and volume of the base.

The volume of acid left will react with the alkali

N1 x 50 ml = $\frac{N}{10}$  x 20ml

N1 = $\frac{N}{25}$

Volume of acid reacted with the divalent metal(V4)= Total volume of acid - Volume of acid left

V4 = V1 - V2

V4 = 250 - 50

     = 200ml

     = 0.2 l                                  (1l = 1000ml)

We know that gram equivalent (g eq) is given as

g eq = $\frac{Weight}{Equivalent Weight}$  = Normality x Volume(in l)

$\frac{0.18}{Equivalent Weight}$ =  $\frac{N}{25}$ x V4

$\frac{0.18}{Equivalent Weight}$ =  $\frac{N}{25}$ x 0.2

Equivalent weight = $\frac{45}{2}$g

Since the metal is divalent its valency will be 2

Equivalent weight = $\frac{Atomic Mass}{Valency}$

$\frac{45}{2}$ = $\frac{Atomic Mass}{2}$

Atomic mass = 45g

Therefore, the atomic mass of the element will be 45g.