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When 2.46 g of a hydrated salt (MSO(4)xH...

When 2.46 g of a hydrated salt `(MSO_(4)xH_(2)O)` is completely dehydrated, 1.20 g of anhydrous salt is obtained. If the molecular weight of anhydrous salt is 120 g `mol^(-1)`, what is the value of x ?

A

2

B

4

C

5

D

7

Text Solution

Verified by Experts

The correct Answer is:
D
`{:(MSO_(4) .xH_(2)O overset(Delta),rarr,MSO_(4),+,H_(2)O),(2.46g,,120g,,x g):}`
Molecular weight of `MSO_(4) = 120g//mol`
Molecular weight of `MSO_(4).xH_(2)O`
`= 120g + x xx 18g`
(120g + x.18 g) of `MSO_(4).xH_(2)O` on complete
dehydration gives 120 g of `MSO_(4)`
1g gives `= (120)/(120 + 18.x)`
Then, 2.46 g of `MSO_(4) .xH_(2)O` gives `(120 xx 2.4 g)/(120 + 18g)` which is equal to 120g
`therefore (120 xx 2.46)/(120 + 18 x) = 1.20`
`295.2 = 1.20 xx 120 + 1.20 xx 18 x`
`295.2 = 144 + 2.16 x`
`295.2 - 144 = 21.6 x`
`x = (151.2)/(21.6) = 7`
x = 7
`therefore` The value of x = 7
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