What volume of 98% sulphuric acid (in ml) should be mixed with water to obtain 200 mL of 15% solution of sulphuric acid by weight ? Given density of H 2 O = 1.00 g cm –3 , sulphuric acid (98%) = 1.88 g cm –3 and sulphuric acid (15%) = 1.12 g cm –3 .
Text Solution
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It is a case of dilution , simplest way is to determine normality of the 98% and 15% H 2 SO 4 .
(i) For 98% H 2 SO 4 : 100 g H 2 SO 4 solution = 
cm 3 solution has H 2 SO 4 = 98 g
N 1 (98%) = 
= 
= 37.6 N
where w 1 is the weight of solute of equivalent weight E 1 in V mL solution
(ii) For 15% H 2 SO 4 : 100 g H 2 SO 4 solution = 
cm 3 has H 2 SO 4 = 15 g
N 2 (15%) = 
= 
= 3.43
Using N 1 V 1 = N 2 V 2
37.6 × V 1 = 3.43 × 200
V 1 = 18.2 mL
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