When 1 x 10-3 mol of the chloride of an element Y was completely hydrolysed, it was found that the resulting solution required 20 mL of 0.1 M aqueous silver nitrate for complete precipitation of the chloride ion. Element Y could be
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a
Aluminium
b
Phosphorus
c
Silicon
d
Sulphur
answer is D.
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Detailed Solution
YCln+nH2O⟶nHCl+Y(OH)n(n= valency of Y)nHCl+nAgNO3⟶nAgCl=1n=1×10−320×0.1×10−3 Molaity × Volume 1000=Moln=2( others have n>2)