- Organic Chemistry
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- Alkyl Halides, Alcohols and ethers
- Amines and other nitrogen compounds
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- Chemistry in daily life
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# 214 - Chemistry Questions Answers

Two oxides of a metal contain 72.4% and 70% of metal respectively. If formula of 2nd oxide is M_{2}O_{3}, find that of the first?

Sir the answer is M_{3}O_{4}. 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 E^{0} VALUES OF FOLLOWING REDUCTn Rxn R GIVEN

Fe 3+ (aq ) + e ----> Fe3+ (aq) E^{0}= 0.771V

Fe2+(aq) + 2e----> Fe (s) E^{0} =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 = n_{1}FE_{1} + n_{2}FE_{2} = 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

Fe^{3+}(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

2NH_{3} (l) ------> NH_{4}+ (am) + NH_{2}^{-} (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 -32.7^{0} C ( NORMAL BOILING B.P OF NH3 (l) IS -33.4^{0} 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 conc^{n} , &the right half cell initially had the same conc^{n} 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 log_{10}Ksp would have been ?? ( ANS -9.804)

(AMU 2012)

**Asked By: SARIKA SHARMA**

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**Joshi sir comment**

E_{cell} = E_{C} - E_{A }- (0.0591/1) logs

on taking the values 0.29 = -0.0591 logs

or logs = -0.29/0.0591 = -4.907

now K_{sp} = s^{2}

so logK_{sp} = 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 = K_{f}m 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 P_{1}V_{1}/T_{1} = P_{2}V_{2}/T_{2}

after getting the value of V_{2} , we can say that V_{2} volume contains 10.1 gm so 22.4 litre will contain 10.1*22.4/V_{2}

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^{0 } 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, Li_{2}o.

**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