Equivalent

c

b

it forms BaSo4 and H2O2

d

Option a is Cr (V)

a

b

d

a

b

just equate equivalents

w/18.67 = 162.52 / (18.67 + 35.5)

Here, there isnt any change in OS

so here, k2cro4 behaves as a salt

hence N factor = 2

b

It takes 2H+ to replace the extra Ca2+

hence n favtor = 2

c

As2S3 gets 28 electrons removed to form the products

2 electron exchanged

Thus n factor of H2so4 = 2/ mols in balanced eqn

= 1

thus ans is a

b

ORR

Equivalent weight of carbonate = M + 60/2

= M + 30

Equivalent weight of metal oxide = M +8

So M+8/M+30 = 0.6

this also gives 25

b

using dulong petit, Molar mass is roughly 75.29

So valency becomes M/E = roughly 3 = 3 (valency must be integer)

so actual mass = 25* 3

f is just percentage purity ig

ans is b

d

just equate equivalents

b

a

a

25* 0.107 * n = 23.1 * 0.115

Same with part 2, just equate equivalents again

c

a

A and C