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Calculate the molality of a sulfuric acid solution of specific gravity 1.2 containing 27% h2so4 by weight.

Calculate the molality of a sulfuric acid solution of specific gravity 1.2 containing 27% h2so4 by weight.

Grade:11

3 Answers

varun
304 Points
6 years ago
HELLO ADITI,
Let us assume that volume of a solvent = 1 litre
Therefore mass of 1 litre of solvent(water) = 1kg
And specific gravity = 1.2g/L
So,
molality = percentage purity X specific gravity X volume of solvent
                 ----------------------------------------------------------------------------------
                                     Molar mass X mass of solvent
                = 0.27 X 1.2g/L X 1L
                ----------------------------
                          98g/mol X 1kg
so
molality = 0.0033m H2SO4
THANK YOU
               
Nishu
11 Points
6 years ago
Given that the solute(H2SO4) is 27% then the solvent will be 73%Molality= n of h2so4/ weight of solvent in kg (27g×1000)÷(98×72) = 3.8m Here 98 is the molar mass of H2SO4 and 1000 is used to convert 72g to 72kg
Ravneet
104 Points
5 years ago
27% H2SO4 by mass means 1000g of solution contains 270g of H2SO4
Number of moles of H2SO4 = 270g / (98 g/mol) = 2.75 mol
Mass of solvent = 1000g - 270g = 730g = 0.73 kg
Molality = Number of moles of solutes / Mass of solvent in kg
Molality = 2.75 / 0.73 ≈ 3.8m
 

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