12.2 g of benzoic acid (M=122 gmol−1) in 100 g water has an elevatio – InterviewSolution

1.	12.2 g of benzoic acid (M=122 gmol−1) in 100 g water has an elevation in boiling point of 0.27. Kb=0.54 Kkgmol−1. If polymerisation is a 100 % complete, the number of molecules of benzoic acid in an associated state is:
Answer» 12.2 g of benzoic acid $(M = 122 g m o l^{- 1})$ in 100 g water has an elevation in boiling point of 0.27. $K_{b} = 0.54 K k g m o l^{- 1} .$ If polymerisation is a 100 % complete, the number of molecules of benzoic acid in an associated state is:

Discussion

No Comment Found

Related InterviewSolutions

33 If the molality of the aqueous solution is 0.111 then the mole fraction of solute will be
Water lost through transpiration is in the form of water vapor. Hence, it is pure water.
when one mol of oxalic acid is treated wih excess NAOH in dil aq solution ,108 KJ of heat liberated ,then the enthalpy of ionisation of the oxalic acid is (H+OH=H2O)△ H=57.3KJ/mol
what is plorized by refrection?
42. in an atom 2K,8L,8M,2N electrons are present if m=0,s=0.5, then find number of such electron a.6 b.2 c.8d.16
For the reaction, PCl3 (g) + Cl2 (g) = PCl5 (g) , ∆H = -x KJ If the ∆ Hf°PCl3 is -y KJ, what is ∆Hf°PCl5?
63. The edge length of FCC cell is 508pm.If radius of action is 110pm ,what will be radius of anion and also describes the two dimensional lattices and three dimensional primitive unit cells
45. Assertion:In dichromate ion, Size of Cr atom is nearly Tl Reason: All Cr-O bond length in dichromate ion are equal. A)A and R are correct, and R is the correct explaination of A b)A and R are correct but Ris not a correct explaination of A c)A is correct but R is not correct d)Both A and R are not correct
1 mole of a gas is changed from its initial value (15L,2atm) to final state (4L,10atm) respectively. If the change can be represented by a straight line in P-V curve. Calculate the maximum temperature attained by the gas.
The angular momentum of the allowed orbits in a H− like species is integral multiple of: