Question

18 g of glucose is added to 178.2g of water. the vapour pressure of water for this solution

Answer 1

U have not given initial value of vapour pressure of water..so pls put that and solve...rest i have done

Hope it helps
Pls mark brainliest ..

Answer 2

Hey !!

18 g of glucose (C₆H₁₂O₆) = 18 / 180 mole = 0.1 mole

178.2 g water (H₂O) = 178.2 / 18 mole = 9.9 mole

∴ Mole fraction of water in the solution

           = 9.9 / 0.1 + 9.9 = 9.9 / 10

    = 0.99          

Vapour pressure of water in the solution

       = Mole fraction of water in the solution  × Vapour pressure of pure water

 = 0.99 × 760 torr  

   = 752.4 torr    

HOPE IT HELPS YOU !!