JEE MainChemistryModeratePhysical Chemistry

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 H2O = 1.00 g cm^-3, sulphuric acid (98 ) = 1.88 g
cm-3 and sulphuric acid (15 ) = 1.12 g cm-3.

Show answer and solution

It is a case of dilution, simplest way is to determine
normality of the 98 and 15 H2SO4.
(i) For 98 H2SO4: 100 g H2SO4 solution

= \frac{100}{1.88}{cm}^{3}

solution has H2SO4 = 98 g

N_{1}(98\%) = \frac{1000w_{1}}{E_{1}V} = \frac{1000 \times 98}{49 \times \frac{100}{1.88}} = 37.6N

where w1 is the weight of solute of equivalent weight E1 in V mL
solution
(ii) For 15 H2SO4: 100 g H2SO4 solution

= \frac{100}{1.12}{cm}^{3}

has H2SO4 = 15 g

N_{2}(15\%) = \frac{1000w_{1}}{E_{1}V} = \frac{1000 \times 15}{49 \times \frac{100}{1.12}} = 3.43

Using

N_{1}V_{1} = N_{2}V_{2}

37.6 \times V_{1} = 3.43 \times 200

V_{1} = 18.2mL

Source: Grass-G / CH-23 Chemistry Paper 31 Dec. FST.docx

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