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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 |
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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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