Physics, asked by sdominicgriffith61, 9 months ago

.161 mol of non-volatile, non-electrolyte solute is
dissolved in 1000 g water at 298 K, the vapour
pressure of solution is (vapour pressure of pure water
at 298 K is 23.51 mm Hg)​

Answers

Answered by knjroopa
0

Explanation:

Given 0.161 mol of non-volatile, non-electrolyte solute is dissolved in 1000 g water at 298 K, the vapour  pressure of solution is (vapour pressure of pure water  at 298 K is 23.51 mm Hg)

  • Now number of moles of solute n = 0.161 mol
  • Number of moles of solvent N = 1000 g / 18 g / mol (since molecular weight of water = 18 g/mol)
  •                                                 = 55.55 mol
  • Mole fraction will be n/N (since n < N)
  •                                  = 0.161 / 55.55
  •                                 = 0.002898
  • Given Po = 23.51 mm Hg
  • We have mole fraction of solute = Po – Ps / Po
  •                                       0.002898 = 23.51 – Ps / 23.51
  •                                      23.51 – Ps = 23.51 x 0.002898
  •                                      23.51 – Ps = 0.06813
  •                                   Or Ps = 23.441 mm Hg
  •     Therefore the vapor pressure of solution is 23.44 mm Hg

Reference link will be

https://brainly.in/question/20520880

Answered by shahanaaz90
1

Answer:

22.441mmHG will be the correct answer hope it helps you

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