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Chemistry >> Physical Chemistry >> Solutions Medical Exam

smiley WHICH OF THE FOLLOWING HAS LOWEST BOILING PT.  (HARY. PMT 2003)

1)  NaCl

2) CuCl

3)  CuCI2

4) CsCl

SIR ANSWER GIVEN IS(3)  . BUT AS THE NO. OF IONS IS MAX IN CuCl2 SO IT SHOULD HAVE MAX BP . THEN WHY ITS BP WILL BE MIN. ??

Asked By: SARIKA SHARMA
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Solution by Joshi sir

 

The only possibility is that the dissociation of the salt CuCl2 is low yet NaCl and CsCl are strongly dissociable salts. But CuCl should be less dissociable than CuCl2 . On this basis answer should be CuCl.

It is also probable that the question will be highest boiling point. 

 
Chemistry >> Organic Chemistry >> General Mechanism in organic compounds IIT JEE

explain the mechanism when alkenes are treated with cold KMnO4 ?

Asked By: MOHIT SAINI
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Joshi sir comment

Chemistry >> General Chemistry >> Mole Concept and Stoicheometry IIT JEE

Sir the formula for converting volume in STP condition is given in my book if volume is not given in STP condition. But my problem is that I don't know how to use it . please explain me this formula and I request you to give me Ques. related to this formula.

Asked By: HEMA RANI
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Joshi sir comment

Formula for converting volume at temperature T to STP is

P1V1/T1 = P2V2/T2

Chemistry >> General Chemistry >> Mole Concept and Stoicheometry IIT JEE

Two oxides of a metal contain 72.4% and 70% of metal respectively. If formula of 2nd oxide is M2O3, find that of the first?    

Sir the   answer is M3O4. But I don't know the steps for doing this type of question, so pls explain me each step!!! 

Asked By: HEMA RANI
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Joshi sir comment

for second oxide ratio of metal and oxygen = 70:30

let molcular mass of metal = M so 70/30 = 2M/3*16 

or 7/3 = 2M/48 so M = 24*7/3 = 56

now for first oxide ratio of metal and oxygen = 72.4/27.6     (By weight)

so ratio by no. of atoms = (72.4/56)/(27.6/16) = 72.4*16/27.6*56 = 1158.4/1545.6 = 3/4

Chemistry >> General Chemistry >> Redox Reactions and n factor Medical Exam

THE E0 VALUES OF FOLLOWING REDUCTn  Rxn  R GIVEN

Fe 3+ (aq ) + e ----> Fe3+ (aq)  E0= 0.771V

Fe2+(aq)  + 2e----> Fe (s) E0 =0.447V

WHAT WILL BE THE FREE ENERGY CHANGE FOR D Rxn ?

Fe3+(aq) + 3E----> Fe(s)

ans +11.87kJ/MOL.

Asked By: SARIKA SHARMA
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Joshi sir comment

Third reaction is the sum of 1st and 2nd so energy of 3rd will be the sum of energies of 1st and 2nd.

delta G = n1FE1 + n2FE2 = 1F(0.771) + 2F(0.447) = F(0.771+0.894) = 1.665F = 1.665*96368 = 160452.7 J = 38203.03 cal = 38.20303 cal

so total energy of the reaction 

Fe3+(aq) + 3e- -------> Fe(s) is 38.203 cal

and value of n = 3 so per mole energy = 38.203/3 

 

 

Chemistry >> Physical Chemistry >> Chemical and Ionic Equilibrium Medical Exam

QUES)  AMMONIA UNDERGOES SELF DISSOCIATION ACCORDING TO D Rxn

2NH3 (l) ------> NH4+ (am) + NH2-  (am)

WHERE am , STANDS FOR AMMONIATED . WHEN 1 MOL OF NH4Cl IS DISSOLVED IN 1 kg OF LIQUID AMMONIA , THE B.P. AT 760 TORR IS OBSV. AT -32.70 C ( NORMAL BOILING B.P OF NH3 (l) IS -33.40 C )

WHAT CONCLUSn  ARE REACHED ABOUT THE NATURE OF SOLn??

( ANS NH4Cl IS COMPLETELY DISSOCIATED IN NH3)

Asked By: SARIKA SHARMA
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Joshi sir comment

boiling point elevation = (-32.7) - (-33.4) = 0.7

now according to the given data 

boiling point elevation = 1000*K(ammonia)*1/1 = 1000*0.00035 = 0.35

so i = 0.7/0.35 = 2

so 100 % dissociation

Chemistry >> Physical Chemistry >> Electrochemistry Medical Exam

consider two half cells based on the Rxn

Ag+(aq) +e- ----> Ag(s)

the left half cell contain Ag+ ions at unit concn , &the right half cell initially had the same concn of Ag+ ions , but just enough NaCl (aq) had been added to completely precipitate the Ag+(aq) as AgCl . if emf ofthe cell is 0.29 V , then log10Ksp would have been ??                                                                                                                                    ( ANS -9.804)

(AMU 2012)

Asked By: SARIKA SHARMA
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Joshi sir comment

Ecell = EC - EA - (0.0591/1) logs

on taking the values 0.29 = -0.0591 logs

or  logs = -0.29/0.0591 = -4.907

now Ksp = s2

so logKsp = 2logs = -9.8

 

 

Chemistry >> Physical Chemistry >> Solutions IIT JEE

What approximate proportions by volume of water( d=1g/cc) and ethylene glycol C2H6O2 (d=1.12g/cc) must be mixed to ensure protection of an automobile cooling system to -10 C?

Asked By: AMIT DAS
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Joshi sir comment

we know that  dT = Kfm so m = 10/1.86

m means moles of ethelene glycol in 1000 gm water

so 1000 gm water contains (10/1.86) * 62 gm ethylene glycol

so 1000 cc water contains [(10/1.86)* 62] / 1.12 cc

now calculate

Chemistry >> Physical Chemistry >> Solutions IIT JEE

10.1 g of a volatile liquid occupies a volume of 4 litres when vaporised at 100 C and 70 cm pressure. what would be the osmotic pressure of a 2% ( grams per 100 cc) solution of this substance  at 0 C?

Asked By: AMIT DAS
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Joshi sir comment

10.1 g of a volatile liquid occupies a volume of 4 litres when vaporised at 100 C and 70 cm pressure.

first convert this volume to STP by using the formula P1V1/T1 = P2V2/T2

after getting the value of  V2 , we can say that V2 volume contains 10.1 gm so 22.4 litre will contain 10.1*22.4/V2

thus we will get molar mass

now use πV = (w/M) RT

V = 0.1 litre, w = 2 gms, M = calculated above, T = 273 K

Chemistry >> Physical Chemistry >> Solutions IIT JEE

The vapour pressure of pure benzene is 22 mm & that of pure toulene is 75 mm at 200  C . what is the composition of the solution of these two components that has a vapour pressure of 50 mm at this temperaure? what is the composition of vapour in equilibrium with this solution?

Asked By: AMIT DAS
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Joshi sir comment

x22+(1-x)75 = 50 

so 22x+75-75x = 50

so 53x = 25 so x = 25/53

so ratio of there amounts (in mole) = 25/53 : 28//53    =  25:28

now ratio in vapour form can be obtained by this method

v. p. of pure benzene = 25/53 * 22 = 550/53

and v. p. of pure toluene = 28/53 * 75 = 2100/53

so ratio = (550/53)/(2100/53) = 550/2100 = 55/210 = 11/42

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