Home
Class 12
CHEMISTRY
Mixture of volatile components A and B h...

Mixture of volatile components A and B has a total vapour pressure (in torr)p=`254-119 x_(A)`is where`x_(A)`mole fraction of A in mixture .Hence`P_(A)^@`and `P_(B)^@`are(in torr)

Promotional Banner

Similar Questions

Explore conceptually related problems

Mixture of voltaile omponents A and B has total vapoure pressure (in torr) : p=254-135 x_(A) where x_(A) is mole fraction of A in mixture hence P_(A)^(@) and P_(B)^(@) are (in torr)

Mixture of volatile components A and B has total vapour pressure (in torr) : P=254-119X_(A) Where X_(A) is mole fraction of A in mixture . Hence, P_(A)^(0)& P_(B)^(0) are (in torr):

A mixture of volatile components A and B has total vapour pressure (in torr) P=254-119 chi_(A) where chi_(A) is the mole fraction of Ain mixture. Hence, P_(A)^@ and P_(B)^@ are (torr)

Mixture of volatile components A and B has total pressure ( in Torr ) p=265-130x_(A), where X_(A) is mole fraction of A in mixture . Hence P_(A)^(@)+P_(B)^(@)= ( in T o r r).

Mixture of volatile components A and B has total pressure ( in Torr ) p=265-130x_(A), where X_(A) is mole fraction of A in mixture . Hence P_(A)^(@)+P_(B)^(@)= ( in T o r r).

At a given temperature , total vapour pressure (in Torr) of a mixture of volatile components A and B is given by "P"_("total")=120-785"X"_("B") hence, vapour pressure of pure A and B respectively (in Torr) are

Consider a binary mixture of volatile liquides. If at X_(A)=0.4 , the vapour pressure of solution is 580 torr then the mixture could be (p_A^@=300 "torr" ,P_(B)^@=800 "torr") :

Consider a binary mixture of volatile liquides. If at X_(A)=0.4 , the vapour pressure of solution is 580 torr then the mixture could be (p_A^@=300 "torr" ,P_(B)^@=800 "torr") :