# 213 - Chemistry Questions Answers

Q  what vol. of O2 OBTAINED AT ANODE AT STP IF Aq SOLn OF CuSO4 IS ELECTROLYSED BY PASSG 100A CURRENT FOR 60 MIN? ANS 20.89L

Joshi sir comment

Total charge flown in the circuit = 100*60*60 = 360000 coulomb

so no of faradays of electricity = 360000/96500 = 3600/965 = 3.73 F

so amount of O2 = 3.73*5.6 = 20.89 litre

Here 5.6 litre is eq. volume corresponding to 8 gm. of O2

Q ONE LITRE MIXTURE OF CO &CO2 IS PASSED THRU RED HOT CHARCOAL IN A TUBE THE NEW VOL . BECOMES 1.2L WHAT % OF CO IN MIXTURE INITIALLY?

Q AN Aq SOLn CONTAINING 100 g OF DISSOLVED MgSO4 IS FED TO A CRYSTALLIZER WHERE 80% SALT CRY. OUT AS MgSO4.6H2O CRYSTAL HOW MANY gm OF HEXAHYDRATE SALT CRY. ARE OBTAINED FROM CRYSTALLIZER=?

ELECTROCHEM

Fe3+  +  e- -----> Fe2+ = 0.77V

Fe --------.>  Fe3+  + 3e-     E = +0.04V

WHAT IS VALUE OF E* for Fe2+  +2e--------> Fe ?

Joshi sir comment

CO2 + C (charcoal) ------->  2CO

let us consider that initially volume of CO2 is x litre and that of  CO is (1-x) litre

so after the above written process net CO in final = 2x+1-x = 1.2      given

on solving x = 0.2 and 1-x = 0.8   now calculate %

MgSO4  -------------->   MgSO4 .6H2O

it means 120 gm (mw of compound) gives 228 gm MgSO4 . 6H2

so 80 gm will give 228*80/120     solve it

here 80 is used for 80% of 100 gm.

Fe+++  +  e-  ----->  Fe++                  E=  +0.77 V

Fe                ----->  Fe+++  + 3e-       E = +0.04 V

Fe                ----->  Fe++  + 2e-

so E for the new reaction = [1F(0.77) + 3F(0.04)]/2F = 0.89/2 = 0.445

SIR , I WANT TO KNOW THAT IN THE EQUILM QUES WHY YOU TAKE MOL FRACT FOR CH3COCH3=1-1/6-1/6.WHY YOU SUB FROM 1.

Joshi sir comment

If we multiply mole fraction to pressure, we get the pressure exerted by the component so it is a good way to get answer. and 1 represents net sum of mole fractions of all the components. here first and second 1/6 stands for mole fraction of ethane and carbon monooxide.

Q In NAcL type str of A+ &B- ,2A ATOMS ARE REMOVED FROM EDGES &2B ATOMS ARE REMOVED FROM CORNER. FORMULA=?  ANS =A14B15 . HOW?

Joshi sir comment

In NaCl like structure B covers all the corners and face centre and A covers all the edge centres and body centre

Net contribution of B = 8*1/8 + 6*1/2 = 4

Net contribution of A = 12*1/4 + 1*1 = 4

now 2 A ATOMS ARE REMOVED FROM EDGES so remaining contribution = 10*1/4 + 1*1 = 7/2

and 2 B ATOMS ARE REMOVED FROM CORNE so remainnig contribution = 6*1/8 + 6*1/2 = 15/4

so formula will be A7/2B15/4  or A14B15

Q CH3COCH3 (g) eqm CH3CH3 +CO

initial pressure of CH3COCH is 300 mm when eqm is setup mole fraction of CO(g) = 1/6 hence total pressure will be =?

ANS 360mm of hg .SOL =?

Joshi sir comment

At equilibrium mole fraction of CO = 1/6, that of CH3CH= 1/6

so mole fraction of CH3COCH3  = 1- 1/6 - 1/6 = 2/3

let the total pressure at equilibrium = p

so pressure of CH3CH3 and CO will be p/6 and p/6

and that of CH3COCH= 2p/3

now consider the following dissociation data

CH3COCH3 ----> CH3CH3 + CO

300 mm                        0                      0                            at starting

300-x  mm                    x                       x                            at  equilibrium

here x = p/6  (1)  and 300-x = 2p/3   (2)                                         add these equations and get the value of p

sir, can please write atopic on amins for  reactions?

Joshi sir comment

Yes

very soon

Sir can you please post a topic on ionic equilibrium ( especially for numerical solving).

Solution by Joshi sir

YES

very soon

Structure of Na2S4O6

Joshi sir comment

A decapeptide (Mol.wt 796) on complete hydrolysis gives glycine(Mol.wt 75) , alanine and phenylalanine.  Glycine contributes 47 % to the total weight  of the hydrolysed products . The number of glycine units present in the decapeptide is

Joshi sir comment

47 % of (796+162) = 958*47/100 = 450 approx

so no of glycene units = 450/75 = 6 approx

here 162 is taken for the weight of 9 water molecules required for breaking the decapeptides.

The change in standard gibbs free energy of given cell at NTP

Zn I Zn(2+) II OH- I Ag2O I Ag         Eo =1.104V