The density (in g mL-1) of a 3.6 M sulphuric acid solution that is 29 % H2SO4 (Molar mass = 98 g mol-1) by mass will be :

Question

The density (in g mL-1) of a 3.6 M sulphuric acid solution that is 29 % H2SO4 (Molar mass = 98 g mol-1) by mass will be : (A) 1.64 (B) 1.88 (C) 1.22 (D) 1.45

Tardigrade Admin · Accepted Answer

Let the density of solution be 'd' Molarity of solution given = 3.6 i.e., 1 litre of solution contains 3.6 moles of H2SO4 or 1 litre of solution contain 3.6 × 98 g of H2SO4 Since, the solution contains 29 % by mass. 100 g solution contains 29 g H2SO4 (100/d)mL solution contains 29 g of H2SO4 1000 mL solution contains 3.6 × 98 g of H2SO4 ∴ 3.6 × 98 = (29 × d/100) × 1000 d = 1.22

Q. The density (in $g m L^{- 1}$ ) of a $3.6 M$ sulphuric acid solution that is $29% H_{2} S O_{4}$ (Molar mass $= 98 g m o l^{- 1})$ by mass will be :

$1.64$

$1.88$

$1.22$

$1.45$

Q. The density (in gmL−1) of a 3.6M sulphuric acid solution that is 29%H2​SO4​ (Molar mass =98gmol−1) by mass will be :

1.64

1.88

1.22

1.45

Q. The density (in $g m L^{- 1}$ ) of a $3.6 M$ sulphuric acid solution that is $29% H_{2} S O_{4}$ (Molar mass $= 98 g m o l^{- 1})$ by mass will be :

$1.64$

$1.88$

$1.22$

$1.45$