KVPY SA Exam 2014 Question Paper

Question : 73 of 80

Marks: +1, -0

Complete reaction of 2.0 g of calcium (at. wt. = 40) with excess HCl produces 1.125 L of H

_2

gas. Complete reaction of the same quantity of another metal “ M ” with excess HCl produces 1.85 L of H

_2

gas under identical conditions. The equivalent weight of “ M ” is closest to

Solution:

(eq)_\Ca/(eq)_\M=(eq)_{H_2 ,\released}/ (eq)_{H_2 ,\released}

(2/20)/(2/x)={1.125/{(\eq.\vol.)}_{H_2}/{1.85/{(\eq.\vol.)}_{H_2}}

{∵eq wt of M x }

∴ x/20=1.125/1.85

x=1.125/1.85×20≈12

Go to Question:

Prev Question

Next Question