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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 isA. 4B. 5C. 6D. 8 |
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Answer» Correct Answer - 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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