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Two liquids A and B form ideal solution....

Two liquids A and B form ideal solution. At `300 K`, the vapour pressure of a solution containing 1 mole of A and 3 moles of B is `550 mm` of Hg. At the same temperature, if one more mole of B is added to this solution, the vapour pressure of the solution increases by `10 mm` of Hg. Determine the vapour pressure of a and B in their pure states.

A

`p_A^@`=600 mm Hg and `p_B^@`=400 mm Hg

B

`p_A^@`=550 mm Hg and `p_B^@`=560 mm Hg

C

`p_A^@`=450 mm Hg and `p_B^@`=650 mm Hg

D

`p_A^@`=400 mm Hg and `p_B^@`=600 mm Hg

Text Solution

Verified by Experts

The correct Answer is:
D
Vapour pressure of solution containing 1 mole of A + 3 moles of B =550 mm Hg Vapour pressure of solution containing 1 mole of A + 4 moles of B = (550+10) = 560 mm Hg
`P_"Total"=p_A^(@) xx x_A + p_B^(@) xx x_B`
or 550 =`p_A^(@) xx x_A + p_B^(@) xx x_B`
`=p_A^@ xx 1/4 + p_B^@xx3/4[because x_A=1/(1+3)=1/4, x_B=3/(1+3)=3/4]`
`550=p_A^@/4+3/4xxp_B^@` or 2200=`p_A^@+3p_B^@`...(i)
Again , we have
560=`p_A^@xx1/5+p_B^@xx4/5 (because x_A=1/(1+4)=1/5, x_B=4/(1+4)=4/5)`
2800=`p_A^@+4p_B^@`...(ii)
Solving equations (i) and (ii) , we get
`p_B^@`=600 mm Hg , `p_A^@`=400 mm Hg
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