The volumes of HClN4 and HClN10 required to make 1 litre of HClN6 are [Kerala PMT 2004] (a) 0.75 litre of 10 N HCl and 0.25 litre of 4 N HCl (b) 0.25 litre of 4 N HCl and 0.75 litre of 10 N HCl (c) 0.67 litre of 4 N HCl and 0.33 litre of 10 N HCl (d) 0.80 litre of 4 N HCl and 0.20 litre of 10 N HCl (e) 0.50 litre of 4 N HCl and 0.50 litre of 10 N HCl
Correct option is C) Let N 1 and V 1 be normality and volume of 4NHCl and N 2 be normality and volume of 10NHCl respectively. As volume to make is 1L and Normality to make is 6N, so V 1 +V 2 =V As we know N 1 V 1 +N 2 V 2 =NV Calculation to determine V 1 and V 2 : Substituting the values in N 1 V 1 +N 2 V 2 =NV equation 4×V 1 +10×(1−V 1 )=6×1 ( As V 2 =1−V 1 )so V 1 =2/3 and V 2 =1/3 V 1 =2/3=0..67 litre and V 2 =0.33 litre Thus, answer is 0.67 litre of 4NHCl and 0.33 litre of 10NHCl. Hence, the correct option is C
