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Moles of Na2SO4 to be dissolved in 12 mole water to lower its vapour pressure by 10 mm Hg at a temperature at which vapour pressure of pure water is 50 mm is:
- a)1.5 mole.
- b)2 mole.
- c)1 mole.
- d)3 mole.
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
Moles of Na2SO4 to be dissolved in 12 mole water to lower its vapour p...
Given data:
Vapour pressure of pure water = 50 mm Hg
Vapour pressure of water after dissolving Na2SO4 = 50 - 10 = 40 mm Hg
To find: Moles of Na2SO4 required to lower the vapour pressure of 12 moles of water by 10 mm Hg.
We can use Raoult's law to solve the problem.
According to Raoult's law, the lowering of vapour pressure (ΔP) is directly proportional to the mole fraction (x) of the solute in the solution.
ΔP = P°solvent × x
where P°solvent is the vapour pressure of the pure solvent and x is the mole fraction of the solute.
Let the number of moles of Na2SO4 required to lower the vapour pressure of 12 moles of water by 10 mm Hg be n.
Mole fraction of Na2SO4 in the solution = n / (n + 12)
According to Raoult's law,
ΔP = P°water × xNa2SO4
Substituting the given values,
10 = 50 × n / (n + 12)
Solving for n, we get n = 1 mole.
Therefore, the number of moles of Na2SO4 required to lower the vapour pressure of 12 moles of water by 10 mm Hg is 1 mole. Hence, option (c) is the correct answer.
Vapour pressure of pure water = 50 mm Hg
Vapour pressure of water after dissolving Na2SO4 = 50 - 10 = 40 mm Hg
To find: Moles of Na2SO4 required to lower the vapour pressure of 12 moles of water by 10 mm Hg.
We can use Raoult's law to solve the problem.
According to Raoult's law, the lowering of vapour pressure (ΔP) is directly proportional to the mole fraction (x) of the solute in the solution.
ΔP = P°solvent × x
where P°solvent is the vapour pressure of the pure solvent and x is the mole fraction of the solute.
Let the number of moles of Na2SO4 required to lower the vapour pressure of 12 moles of water by 10 mm Hg be n.
Mole fraction of Na2SO4 in the solution = n / (n + 12)
According to Raoult's law,
ΔP = P°water × xNa2SO4
Substituting the given values,
10 = 50 × n / (n + 12)
Solving for n, we get n = 1 mole.
Therefore, the number of moles of Na2SO4 required to lower the vapour pressure of 12 moles of water by 10 mm Hg is 1 mole. Hence, option (c) is the correct answer.
Community Answer
Moles of Na2SO4 to be dissolved in 12 mole water to lower its vapour p...
I think ans is 3 mole and option D is correct
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Moles of Na2SO4 to be dissolved in 12 mole water to lower its vapour pressure by 10 mm Hg at a temperature at which vapour pressure of pure water is 50 mm is:a)1.5 mole.b)2 mole.c)1 mole.d)3 mole.Correct answer is option 'C'. Can you explain this answer?
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