Question

0.8 gm of divalent metal was dissolved in 100 ml of 1.28N HCl and the solution was diluted to 200ml. then 50 ml of the solution required 54.6ml of the solution required 54.6ml of 0.22N NaOH for complete neutralization. Find the atomic weight of metal

Answer 1

Answer:

33 g/mol

Explanation:

Here's the explanation: https://study.webmatrices.com/d/23

Answer 2

Given:

0.8g of divalent metal is dissolved in 100ml of 1.28N HCl the solution was diluted to 200ml, 50ml of the solution required 54.6ml of the solution required 54.6ml of 0.22N NaOH for complete neutralization.

To find:

Atomic weight of divalent metal

Solution:

In a solution of 100ml of 1.28N amount of HCl present is = $\left {\frac{{36.5 \cdot 100 \cdot 1.28}}{{1000 }} \right$ = 4.672g

Let, the normality of diluted acid is Nₐ, then

Nₐ × 50ml = 0.22N × 54.6ml

Nₐ = 0.240N

In 200ml diluted acid solution amount of HCl present =$\left {\frac{{36.5 \cdot 200 \cdot 0.24}}{{1000 }} \right$ = 1.752g

therefore, 0.8g of metal consumed amount of HCl = (4.672-1.752) = 2.92g

 2.92g of HCl = 0.8g of divalent metal

so, 36.5g of HCl = $\left {\frac{{0.8 \cdot 36.5 }}{{2.92 }} \right$ of divalent metal

                           = 10g of divalent metal

therefore, equivalent weight of the metal is 10

Atomic weight = equivalent weight × valency

as, metal is divalent therefore valency is 2

so,  Atomic weight = 10×2 amu

                               = 20 amu

therefore, atomic weight of a divalent metal is 20 amu.