Question

• A solution of chloroform (molar mass 119.5 g mol-1) has been prepared by dissolving 2.96 gram of chloroform in 75 gram of benzene. if Kf for benzene is 5.12 K Kg mol-1 and Kb for benzene 2.5 K kg mol-1. and vapour pressure of benzene at 298 K is 0.256 bar. Q1. the vapour pressure of the solution at 298 K will be (a) 0.942 bar (b) 0.429 bar (c) 0.294 bar (d) 0.249 bar Q2. the osmotic pressure of the solution at 298 K will be (a) 0.080 atm, (b) 0.008 atm, (c) 0.0008 atm, (d) 0.800 atm Q3. the freezing point depression and freezing point of the solution will be (a) -1.49 K & 271.66 K (b) -1.42 K & 271.73 K (c) -1.69 K & 271.46 K (d) -1.62 K & 271.53 K Q4. the boiling point elevation and boiling point of the solution will be (a) 0.83 K & 373.98 K (b) 0.38 K & 373.53 K (c) 0.72 K & 373.87 K (d) 0.27 K & 373.42 K Q5. the mass of benzoic acid that should be dissolved in the same amount of benzene to get the same freezing point will be (a) 4.0 g (b) 5.0 g (c) 6.0 g (d) 8.0 g Q6. what is the percentage dissociation of the benzoic acid if it forms dimer in the solution (a) 98.3 % (b) 99.3 % (c) 94.3 % (d) 96.3 %

give all answer ​

Answer 1

Answer:

here is your answer

Explanation:

ΔT

b

=K

p

×molality

⇒80.31−80.10=2.53×

99

M

1.25

×100

⇒0.21=2.56×

99M

1250

⇒M=

99×0.21

2.53×1250

=152gmmol

−1

Hence the molar mass of this compound is 152gmmol

−1