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Ammonia in 0.224 g of a compound Zn(NH(3...

Ammonia in `0.224 g` of a compound `Zn(NH_(3))_(x)Cl_(2)` is neutralised by `30.7 mL` of `0.20 M HCl`. The value of `x` in the formula is

A

4

B

5

C

6

D

8

Text Solution

Verified by Experts

The correct Answer is:
C
`Mw "of" Zn (NH_(3))_(x) Cl_(2) = 65.30 + 17x + 35.4 xx 2`
`= 136.30 + 17 x`
`(17 x + 36.30)g` of compound contains `= x` mole of `NH_(3)`
`0.224 g` compound `= (X)/(17 + 136.30) xx 0.224`
`x` mol of `NH_(3) = x eq "of"NH_(3)`
Eq of `NH_(3) = Eq "of" HCl`
`(0.244 x)/(17x + 136.30) = (30.7 xx 0.2)/(1000) eq "of"HCl`
`x ~~ 6.75 ~~ 6` (as the choice given)
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