Molality of 18M sulphuric acid(density=1.8grams per ml)is??
(a)36 mol per kg
(b)200 mol per kg
(c)500 mol per kg
(d)18 mol per kg
Answer:
molality = mol H2SO4/kg solvent
molarity = mol H2SO4/L solution
Using density,
Mass of the solution = 1.8g/ml * 1000 ml = 1800g
Molecular weight of H2SO4 = 98g
Mass of solute (H2SO4) = 18 mol * 98 g = 1764 g
Mass of solvent = mass of solution - mass of solute
= 1800 g - 1764 g
= 36 g = 0.0036 kg
Molality = mol. H2SO4/ kg of solvent
= 18 mol/0.0036kg
= 500 mol/kg
(e) 1 year in jail
molality = (moles of solute) / (kilograms of solvent)
(molality is denoted by a lower case m)
First, pick an arbitrary amount of solution that will make the calculations easier, say 1.00L. Since the concentration of the solution is 18.0 mol/L and the density is 1.8 g/ml (or 1.8 kg/ L), 1.00 Liter of the solution contains 18.0 mol (1765 g) of H2SO4 and has a mass of 1.8 grams:
1.
Moles of H2SO4 in 1.0L soln= (18.0 mol/L)x(1.00L) = 18 mol
2.
Mass of H2SO4 in 1.0 L soln=(18 mol)x(98.1g/mol) = 1765 g
(molar mass of sulfuric acid is 98.1g/mol)
3.
Mass of 1.0L soln = (1.00L) x (1.8kg/L) = 1.8 kg
4.
Mass of H20 in 1.0L soln =(1.8kg) - (1.765 kg)=0.035 kg H2O
5.
Since 0.035 kg of water has 18 mol of H2SO4 dissolved in it, 1.00 kg of water would have 514 mol dissolved in it:
1.00kg H20 x (18 mol H2SO4 / 0.035 kg H20) =
= 514 mol H2SO4 (kg cancel)
So, finally, sorry this was so long, but the answer closest to the exact calculation is
C. 500 mol H2SO4 per kilogram of water.
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