.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
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
1
Answer:
22.441mmHG will be the correct answer hope it helps you
Similar questions