Vapour pressure of a pure liquid X is 2 atm at 300 K. It is lowered to 1 atm on dissolving 1g of Y i — Solutions and Colligative Properties Chemistry Question
Question
Vapour pressure of a pure liquid X is 2 atm at 300 K. It is lowered to 1 atm on dissolving 1g of Y in 20g of liquid X. If molar mass of X is 200, what is the molar mass of Y?
Answer: A
💡 Solution & Explanation
0 2 1 x x P at m P at m 1 20 200 / ? Y X X Y W g W g M g mole M 0 0 2 1 1 200 2 .20 1 200 200 2 2 20 20 20 / X X Y X X Y X Y Y Y Y P P W M P M W M M M M g mole x x
💬Ask on WhatsApp →
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes