TS EAMCET 11-Sep-2020 Shift 1 Solved Paper

Given,
Vapour pressure of pure water =23mmHg10% of solute A
Molecular weight of A=50
Mass of A=10g
Mass of water =100−10=90g
Number of moles of A=‌

10

50

=0.2
Number of moles of water =‌

90

18

=5
∴ Total number of moles =5+0.2=52
Mole fraction of solvent =‌

3

5.2

ps= mole fraction of solvent ×p∘
=‌

22

760

=0.028atm