the vapur pressure of an ideal olut...
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the vapur pressure of an ideal oluti l and 2 mole of B OTA the sune temperature, if I 5 mole of A&0.5 mole of of AX 0,5 molec non-volatile) are do o latile are added to this solution the vapour pressure of solution increases by 30 torr. What is the value of P . (A) 940 (B) 405 (C) 90 (D) none of these

JEE/Engineering Exams
Chemistry
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Solution Mole fraction of ( B ) is ( X_{B}=frac{2}{2+3}=0.4 ) The mole fraction of ( A ) is ( X_{A}=1-0.4=0.6 ) The expression for the total pressure of solution is ( P=P_{A}^{o} X_{A}+P_{B}^{o} X_{B} ) Substitute values in the above expression ( P=600 ) torr ( =P_{A}^{o} times 0.6+P_{B}^{o} times 0.4 ldots ldots ) (1) 1.5 mole of ( A & 0.5 ) mole of ( B ) (non-volatile) are added Mole fraction of B is ( X_{B}=frac{2.5}{5+2}=0.36 ) The mole fraction of A is ( X_{A}=1-0.36=0.64 ) The expression for the total pressure of solution is ( P=P_{A}^{o} X_{A}+P_{B}^{o} X_{B} ) The total pressure is ( 600+30=630 ) torr Substitute values in the above expression ( P=630 ) torr ( =P_{A}^{o} times 0.64+P_{B}^{o} times 0.36 ) Rearrange equation (2) ( P_{A}^{o}=frac{630-0.36 P_{B}^{o}}{0.64} ) From (1) and (3) ( 600=0.6left(frac{630-0.36 P_{B}^{o}}{0.64}right)+0.4 P_{B}^{o} ) ( 600=590.625-0.3375 P_{B}^{o}+0.4 P_{B}^{o} ) ( 600=590.625+0.0625 P_{B}^{a} ) ( P_{B}^{o}=frac{600-590.625}{0.0625}=150 ) torr
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