Question

when 0.2 moles of NaCl is added to 2 moles of water then find the vapour pressure of the solution if the standard VP of water is 760mm Hg.​

Answer 1

mole of NaCl , n = 0.2 mol

mole of water, N = 2 mol

so, mole fraction of NaCl(solute), $\mu$ = n/(N + n)

= 0.2/(2 + 0.2)

= 0.2/2.2 = 1/11

Let vapor pressure of solution is P

from lowering of vapor pressure formula,

lowering of vapor pressure = mole fraction of solute

or, $\frac{(P_0-P)}{P_0}=\mu$

or, (760 - P)/760 = 1/11

or, 11(760 - P) = 760

or, 11P = 10 × 760

or, P = 7600/11 = 690.9 mm Hg

hence, vapor pressure of solution is 690.9 mm Hg

Answer 2

Answer:

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