Calculate the millimoles of SeO3(2−​) in solution on the basis of following data: 70 ml of 60/M​ solution of KBrO3​ was added to SeO3(2−​) solution. The bromine evolved was removed by boiling and excess of KBrO3​ was back titrated with 12.5 mL of 25/M​solution of NaAsO2​. The reactions are given below. I. SeO3(2−​) +BrO3(−​) + H+→SeO42−​+Br2​+H2​O II. BrO3(−​) +AsO2(−​) + H2​O→Br(−) +AsO4(3−​)+ H+

SOLUTION:

A six co-ordinate complex of formula CrCl3.6H2O has green colour. One litre of O.1M solution of the complex when treated with excess of AgNO3 gave 28.7 g of white precipitate. The formula of the complex is .... (1) [Cr(H2O)6]CI3 (2) [Cr(H2O)5CI]CI2.H2O (3) [Cr(H2O)4CI2]Cl.2H2O (4) [Cr(H2O)3Cl3].3H2O

SOLUTION: 

Given 28.7 g of white ppt obtained is 28.7/147.5= 0.2 mole Agcl precipitate Since 0.1M complexgives 0.2 mole AgCl means 2Cl− are ionisable or two Cl- ion must be present out of coordination sphere so, complex is [Cr(H2​O)5​Cl]Cl2​.H2​O

If N2 gas is bubbled through water at 293 K, how many millimoles of N2 gas would dissolve in 1 litre of water. Assume that N­2exerts a partial pressure of 0.987 bar. Given that Henry’s law constant for N2 at 293 K is 76.84 kbar.

SOLUTION: The solubility of gas is related to its mole fraction in the aqueous solution. The mole fraction of the gas in the solution is calculated by applying Henry's law. Thus,

As 1litre water contains 55.5 mol of it, therefore, if n represents number of moles of N₂ in solution,

One mole of a non-volatile solute is dissolved in two moles of water. The vapour pressure of the solution relative to that of water is

SOLUTION:

Mole fraction of solute in solution Or

                                                                 

The composition of vapour over a binary ideal solution is determined by the composition of the liquid. If XA and YA are the mole-fraction of A in the liquid and vapour, respectively find the value of XA for which YA-XA­ has a minimum. What is the value of the pressure at this composition?

SOLUTION