Calculate the molality of sulphuric acid solution in which mole fraction of water is .85
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Take enough solution to have 1 mol total. That means you have 0.15mol H2SO4 and 0.85 mol H2O. Convert H2O to grams which is 0.85 x (18 g H2O/mol H2O) = approx 15
Then m = mol/kg solvent
m = 0.15/0.015 = ? estimated.
Ans :9.8
To calculate the molality of a solution, you need to know the number of moles of solute dissolved in a given mass of solvent. In this case, the solute is sulphuric acid and the solvent is water. The mole fraction of water is 0.85.
To find the molality, we need to determine the mole fraction of sulphuric acid in the solution, and then convert it to molality. The equation for mole fraction is:
Mole Fraction = Moles of Component A / Total Moles of Solution
Since we know the mole fraction of water is 0.85, we can calculate the mole fraction of sulphuric acid using the equation:
0.85 = Moles of Water / Total Moles of Solution
Let's assume the total moles of solution is 1. So, the moles of water is 0.85.
To convert the mole fraction to molality, we need to know the moles of solute per kilogram of solvent. Since the solvent is water, we can convert the moles of water to kilograms using the molar mass of water, which is approximately 18 g/mol.
Moles of water = 0.85
Molar mass of water = 18 g/mol
Mass of water = Moles of water * Molar mass of water
Mass of water = 0.85 * 18 g
Mass of water = 15.3 g (approximately)
Since the mole fraction relates to the moles of solute per moles of solution, we assume that the moles of sulphuric acid are also 1 to maintain the ratio. So, the moles of sulphuric acid are 1.
Molality = Moles of solute / Mass of solvent (in kilograms)
Molality = 1 mol / 0.0153 kg (since 15.3 g = 0.0153 kg)
Therefore, the molality of the sulphuric acid solution is approximately 65.36 mol/kg.