1.

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. 20B. 50C. 100D. 200

Answer» 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


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