Let the volume be ‘V’ Lt. ∴ Moles of 3 2 .6 0.4 FeCl H O M V V x x ∴ Moles of 3 0.4 Fe V x ∴ Mass of 3 0.4 56 Fe V x x g But the mass required is 600mg i.e. 3 600 10 x g 3 3 3 0.4 56 600 10 600 10 26.78 10 26.78 0.4 56 V V L ml x x x x x x The correct option is (b)
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