Question

Temperature of Chlorobenzene and Bromobenzene in R 0.35 0 bar and 0.500 bar Respectively find the temperature of pollution of ten by mixing 11.2 5 g chlorobenzene with 31.4 g Bromobenzene​

Answer 1

Answer

p0-p

____== w2

p0

p

p 0

p 0 =0.850bar= vapour pressure of pure benzene.

p 0 =0.850bar= vapour pressure of pure benzene.p=0.845bar= vapour pressure of solution.

p 0 =0.850bar= vapour pressure of pure benzene.p=0.845bar= vapour pressure of solution.W

p 0 =0.850bar= vapour pressure of pure benzene.p=0.845bar= vapour pressure of solution.W 1

p 0 =0.850bar= vapour pressure of pure benzene.p=0.845bar= vapour pressure of solution.W 1

p 0 =0.850bar= vapour pressure of pure benzene.p=0.845bar= vapour pressure of solution.W 1 =39g= mass of benzene

p 0 =0.850bar= vapour pressure of pure benzene.p=0.845bar= vapour pressure of solution.W 1 =39g= mass of benzeneW

p 0 =0.850bar= vapour pressure of pure benzene.p=0.845bar= vapour pressure of solution.W 1 =39g= mass of benzeneW 2

p 0 =0.850bar= vapour pressure of pure benzene.p=0.845bar= vapour pressure of solution.W 1 =39g= mass of benzeneW 2

p 0 =0.850bar= vapour pressure of pure benzene.p=0.845bar= vapour pressure of solution.W 1 =39g= mass of benzeneW 2 =78g= mass of solute

p 0 =0.850bar= vapour pressure of pure benzene.p=0.845bar= vapour pressure of solution.W 1 =39g= mass of benzeneW 2 =78g= mass of soluteM

p 0 =0.850bar= vapour pressure of pure benzene.p=0.845bar= vapour pressure of solution.W 1 =39g= mass of benzeneW 2 =78g= mass of soluteM 1

p 0 =0.850bar= vapour pressure of pure benzene.p=0.845bar= vapour pressure of solution.W 1 =39g= mass of benzeneW 2 =78g= mass of soluteM 1

p 0 =0.850bar= vapour pressure of pure benzene.p=0.845bar= vapour pressure of solution.W 1 =39g= mass of benzeneW 2 =78g= mass of soluteM 1 =78g/mol= Molar mass of benzene.

p 0 =0.850bar= vapour pressure of pure benzene.p=0.845bar= vapour pressure of solution.W 1 =39g= mass of benzeneW 2 =78g= mass of soluteM 1 =78g/mol= Molar mass of benzene.M

p 0 =0.850bar= vapour pressure of pure benzene.p=0.845bar= vapour pressure of solution.W 1 =39g= mass of benzeneW 2 =78g= mass of soluteM 1 =78g/mol= Molar mass of benzene.M 2

p 0 =0.850bar= vapour pressure of pure benzene.p=0.845bar= vapour pressure of solution.W 1 =39g= mass of benzeneW 2 =78g= mass of soluteM 1 =78g/mol= Molar mass of benzene.M 2

p 0 =0.850bar= vapour pressure of pure benzene.p=0.845bar= vapour pressure of solution.W 1 =39g= mass of benzeneW 2 =78g= mass of soluteM 1 =78g/mol= Molar mass of benzene.M 2 = Molar mass of solute.

p 0 =0.850bar= vapour pressure of pure benzene.p=0.845bar= vapour pressure of solution.W 1 =39g= mass of benzeneW 2 =78g= mass of soluteM 1 =78g/mol= Molar mass of benzene.M 2 = Molar mass of solute.0.850bar

p 0 =0.850bar= vapour pressure of pure benzene.p=0.845bar= vapour pressure of solution.W 1 =39g= mass of benzeneW 2 =78g= mass of soluteM 1 =78g/mol= Molar mass of benzene.M 2 = Molar mass of solute.0.850bar0.850bar−0.845bar

p 0 =0.850bar= vapour pressure of pure benzene.p=0.845bar= vapour pressure of solution.W 1 =39g= mass of benzeneW 2 =78g= mass of soluteM 1 =78g/mol= Molar mass of benzene.M 2 = Molar mass of solute.0.850bar0.850bar−0.845bar

p 0 =0.850bar= vapour pressure of pure benzene.p=0.845bar= vapour pressure of solution.W 1 =39g= mass of benzeneW 2 =78g= mass of soluteM 1 =78g/mol= Molar mass of benzene.M 2 = Molar mass of solute.0.850bar0.850bar−0.845bar =

p 0 =0.850bar= vapour pressure of pure benzene.p=0.845bar= vapour pressure of solution.W 1 =39g= mass of benzeneW 2 =78g= mass of soluteM 1 =78g/mol= Molar mass of benzene.M 2 = Molar mass of solute.0.850bar0.850bar−0.845bar = 39g×M

p 0 =0.850bar= vapour pressure of pure benzene.p=0.845bar= vapour pressure of solution.W 1 =39g= mass of benzeneW 2 =78g= mass of soluteM 1 =78g/mol= Molar mass of benzene.M 2 = Molar mass of solute.0.850bar0.850bar−0.845bar = 39g×M 2

p 0 =0.850bar= vapour pressure of pure benzene.p=0.845bar= vapour pressure of solution.W 1 =39g= mass of benzeneW 2 =78g= mass of soluteM 1 =78g/mol= Molar mass of benzene.M 2 = Molar mass of solute.0.850bar0.850bar−0.845bar = 39g×M 2

p 0 =0.850bar= vapour pressure of pure benzene.p=0.845bar= vapour pressure of solution.W 1 =39g= mass of benzeneW 2 =78g= mass of soluteM 1 =78g/mol= Molar mass of benzene.M 2 = Molar mass of solute.0.850bar0.850bar−0.845bar = 39g×M 2

p 0 =0.850bar= vapour pressure of pure benzene.p=0.845bar= vapour pressure of solution.W 1 =39g= mass of benzeneW 2 =78g= mass of soluteM 1 =78g/mol= Molar mass of benzene.M 2 = Molar mass of solute.0.850bar0.850bar−0.845bar = 39g×M 2 0.5g×78g/mol

p 0 =0.850bar= vapour pressure of pure benzene.p=0.845bar= vapour pressure of solution.W 1 =39g= mass of benzeneW 2 =78g= mass of soluteM 1 =78g/mol= Molar mass of benzene.M 2 = Molar mass of solute.0.850bar0.850bar−0.845bar = 39g×M 2 0.5g×78g/mol

p 0 =0.850bar= vapour pressure of pure benzene.p=0.845bar= vapour pressure of solution.W 1 =39g= mass of benzeneW 2 =78g= mass of soluteM 1 =78g/mol= Molar mass of benzene.M 2 = Molar mass of solute.0.850bar0.850bar−0.845bar = 39g×M 2 0.5g×78g/mol

Answer 2

Answer:

hi

Explanation:

wruucjhxkxgoxoyxoyzoyzydoydo