An ionic compound ( X + Y ) is doped with controlled amount of ( MY 2 ) . Number of cation vacancies in 1 mol of XY is found to be 6.02 × 1 0 21 . If atomic mass of X, Y and M are 15,10 & 20 a.m.u

Question

An ionic compound ( X + ​ Y ) is doped with controlled amount of ( MY 2 ​ ) . Number of cation vacancies in 1 mol of XY is found to be 6.02 × 1 0 21 . If atomic mass of X, Y and M are 15,10 & 20 a.m.u respectively : (Assume volume remains unchanged) (Given : N A = 6.02 x 10 23 ) If the % decreases in density due to this defect is x, then the value of 10x is :

TestHub · Accepted Answer

Number of moles of cation vacancy ] = 6.02 × 1 0 23 6.02 × 1 0 21 ​ = 1 0 − 2 mole Number of moles of M + 2 ions = 1 0 − 2 mole Number of moles of missing y − ions = 2 × 1 0 − 2 mole % decreases in density ​ = 25 30 × 1 0 − 2 − 20 × 1 0 − 2 ​ × 100% = 0.4% = x % ​ ⇒ 10 x = 4 ​

Chemistry - Solid State Question with Solution | TestHub

Doubts & Discussion