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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)
    3 mole.
  • d)
    1 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 (P°) = 50 mm Hg
Change in vapour pressure (ΔP) = 10 mm Hg
Moles of water (n) = 12 mol

To calculate the moles of Na2SO4 required to lower the vapour pressure of water by 10 mm Hg, we can use the formula:

ΔP/P° = X2
where X2 is the mole fraction of solute (Na2SO4) in the solution.

Rearranging the formula, we get:
X2 = ΔP/P°

Substituting the given values, we get:
X2 = 10/50 = 0.2

Now, we know that the mole fraction of solute is given by:
X2 = moles of solute (Na2SO4)/(moles of solute + moles of solvent)

Since we are dissolving Na2SO4 in water, the solvent is water and the moles of solvent are given as 12 mol. Therefore, we can write:

0.2 = moles of Na2SO4/(moles of Na2SO4 + 12)

On solving, we get:
moles of Na2SO4 = 3

Hence, the correct option is (c) 3 mole.
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Community Answer
Moles of Na2SO4 to be dissolved in 12 mole water to lower its vapour p...
∆p/p=x2
10/50=n2/n2+12
n2=3
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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)3 mole.d)1 mole.Correct answer is option 'C'. Can you explain this answer?
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