Vapour pressure of a pure liquid X is 2 atm at 300 K. It is lowered to 1 atm on dissolving 1 g of Y in 20 g of liquid X. If molar mass of X is 200, wh

Question

Vapour pressure of a pure liquid X is 2 atm at 300 K. It is lowered to 1 atm on dissolving 1 g of Y in 20 g of liquid X. If molar mass of X is 200, wh

Vapour pressure of a pure liquid X is 2 atm at 300 K. It is lowered to 1 atm on dissolving 1 g of Y in 20 g of liquid X. If molar mass of X is 200, what is the molar mass of Y ?
A. 20
B. 50
C. 100
D. 200

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - A
`(p^@-p_A)/p_A^@=n_B/n_A`
`(2-1)/2=W_B/M_BxxM_A/W_A rArr 1/2=1/M_Bxx200/20`
`rArr M_B=(200xx2)/20`=20

5 views Answered 2023-02-05 15:29