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CGP EDU Academic Team
Published on: August 14, 2026
18 g glucosse (C 6 H 12 O 6 ) is added to 178.2 g water. The vapor pressure of water (in torr) for this aqueous solution is.
Text Solution
Verified by ExpertsThe correct answer is:
B
Moles of glucose = 
= 0.1 (Assume that glucose is added at boiling point of water)
Moles of water = 
= 9.9
⇒ n Total = 10
⇒ 
= 
Δ P = 0.01 Pº = 0.01 × 760 = 7.6 torr
P S = 760 – 7.6 = 752.4 torr
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