214 - Chemistry Questions Answers

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

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 

 

 

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

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 - E- (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

 

 

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

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

The vapour pressure of pure benzene is 22 mm & that of pure toulene is 75 mm at 20 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

the max proportion of available volume that can be filled by hard spheres in diamond?

Asked By: SARIKA
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Joshi sir comment

 

In Diamond, there is 8 lattice points per unit cell.  The maximum radius that a hard sphere can have is , r and a are in their usual meaning, Hence the packing fraction is:

in my the previous query , i want to ask the second nearest, third nearest, 4th nearest neighbour atoms/ cordination no. of NaCl, CsCl, ZnS, CaF2, Li2o.

Asked By: SARIKA
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Joshi sir comment

second, third and fourth coordination number means to count the no. of atoms of either Cl or Na, having second, third and fourth closest distance

for NaCl         first coordination number =    6   (6 faces)            distance (a/2)

                 second coordination number =  12   (12 edges)        distance (a/√2)

                       third coordination number =    8   (8 corners)        distance (a√3/2)

                    fourth coordination number =    6    ( 6 centre)          distance (a)

similarly try for others

an element crystallises in fcc lattice having 400pm. calculate the maxm diameter which can be placed in a interstitial sites without disturbing the str

1) 400pm

2) 117.1 pm

3) 224.2 pm

4) 336.2 pm

Asked By: SARIKA SHARMA
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Joshi sir comment

according to the given conditions

along the face of a side 4x = 400√2             or  x = 100√2                 here x is radius of the atom forming lattice

and along an edge 2x+2y = 400                      here y is interstitial radius

so y = 200-x = 200-100√2  = 100 (2-1.414) = 100(0.586) = 58.6 pm

so diameter = 117.2

 

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