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Molecular  Weight  of Organic Acids

1. Silver Salt method

Silver salt of org. acid on combustion gives silver.

RCOOAg —–>Ag

(R= alkyl gr.)

Eq. wt. of org. Acid (RCOOH)= E

Eq. wt. of RCOOAg =E–1+108 = E+107

Eq.Wt. of silver salt / Eq.Wt. of silver = Wt. of silver salt /Wt. of silver

( E+107) /108 =W/w

Mol wt. of acid = Eq. wt × basicity of acid

Q –0.5 gm. of Ag. salt of dibasic acid on combustion give 0.355 gm. of Ag. Find out the  molecular weight of Compd.

W= 0.5, w= 0.355

( E+107) /108 =W/w

( E+107) /108 =0.5/0.355

E =45

Mol wt. of acid = Eq. wt × basicity of acid

=45×2=90 Ans.

Q  –

W= 0.9 g, w= 0.582

Acid is monobasic, Mol. wt.=?

Ans.

( E+107) /108 =W/w

( E+107) /108 =0.9 /0.582

E = 167

Mol. wt. of acid = Eq. wt × basicity of acid

=167 x 1 =167

Ans. Mol .wt. = 167

(II)     Volumetric Method

gm. equi. of Acid = gm. equi. of base

w /E =N V(l)

w=wt of acid,            E=Eq. wt of acid

N= Normality of base          V= Vol. of base in litre

Mol./ wt. = Eq. wt. × basicity

Q — 0.12 gm. of a monobasic acid is neutralised  by 20ml of N/5 NaOH. Calculate the  molecular weight of acid ?

Ans.    w /E =N V (l)

w =0.12 gm. ,N =1/5 =  0.2 , V =20 ml. =0.020 litre

0.12 /E = 0.2 x 0.020

E = 30

Mol wt. = Eq. wt. × basicity

= 30×1=30 Ans.

Q  —  1.575 gm of an org. acid was dissolved in 250 ml of H2O &  20 ml of this Soln requires 16ml of N/8 alkali soln for complete neutralisation. If the basicity is 2 . Find the mol. wt.

For Acid

w=1.575 gm.             N=1/10                       V= 250 ml.=0.250 litre

w /E =N V(l)

1.575 /E =  0.250/10

E =63

Mol . wt = Eq. wt. × 2 = 63×2 = 126 Ans.

III      Molecular weight of organic base-

• Volumetric Method-

gm. eq. of base = gm. eq. of acid

w /E =N V(l)

w= wt. of base                                  N= normality of acid

E= Eq. wt. of base                           V= Vol. of Acid in litre

Mol. wt. =  Eq. wt. × acidity

Q—0.12 gm. of Mono acidic base is completely neutralised by 20 ml of 0.2 N HCl. Calculate the molecular weight of base ?

Ans.  w =0.12 gm. , N=0.2 ,V =20 ml. =0.020 litre

w /E =N V(l)

0.12 /E =0.2 x 0.020

E=30

Mol. wt. =  Eq. wt. × acidity

Mol wt.  =30×1=30 Ans.

Q—-2.65 gm of di acidic base was dissolved in 500 ml of H2O. 20 ml of this soln required 12 ml of N/6 HCl soln. Calculate the eq. wt. &  molecular weight of base.

 N1×20=1/6×12 N1=1/10 for base w=2.65 N=1/10 V=250 ml =0.250 l w/E =NV(l) 2.65 /E =0.250 /10 E =53 Mol. wt. =  Eq. wt. × acidity =53 x 2 Mol.Wt=   106 Ans.
1. b) Platini Chloride Method-

H2PtCl6 —–>  B2H2PtCl6 ——> Pt

Chloroplatinic   Chloro Platinate

Acid                                 Salt

(B is a base)

Eq. wt. of base = E

Wt. of Chloro platinate salt = W gm.

Mol. wt. of Chloro platinate  salt (B2H2PtCl6)= 2E+2+195+35.5×6

= 2E+410

Mol. wt. of Chloro platinate  salt / At.Wt. of Pt =  Wt. of Chloro platinate salt / wt.of Pt

(2E+410) /195 =W/w

Mol. wt. of base = Eq. wt. x Acidity

Q — 0.40 gm platini chloride of a monoacidic  base on combustion gave 0.125 gm of Pt. find out the Mol. wt. of base.

Ans.   W =0.40 , w= 0.125

(2E+410) /195 =W/w

(2E+410) /195 =0.40/0.125

E =107

Mol. wt. of base = Eq. wt. x Acidity

Mol. Wt. = 107 x 1

Mol. Wt. =107

Determination of  Mol. wt. by Physical Methods-

For Volatile compound

1. Victor Mayer’s Method
2. Hofmann’s Method
1. For nonvolatile compound
1. Elevation in boilng point method
2. Depression in freezing point method

Hofmann’s Method-

This method is applied to those substances  which are not stable at their boilng points but which may be volatalize without decomposition under reduced pressure.

Mol. Wt. =( wt. of substance x 22400) / volume of vapours at N.T.P. (ml.)

Victor Mayer’s Method-

Mol. Wt. =( wt. of substance x 22400) / volume of vapours at N.T.P. (ml.)

Volume at N.T.P. can be calculated by Gas eqn

P1 V1/T1 =P2 V2/T2

P, v, T2 are at N.T.P.

P2= 760 mm               V2= ? ml                      T=273 K

Vapour Density =wt.of  vapours or substance /(volume of vapours at  N.T.P. X 0.00009 )

Q —In victor Mayer determination the following observations  have been made. wt. of compd= 0.17 gm.

Vol. of air collected = 34.2 ml., temp. = 150C

Atm.  pressure = 750 m.m., Vapour Pressure of water at 150C= 13 mm

Calculate  the V.D. & Mol. wt. of compd.

Ans.

P1 = 750-13=737 mm

T1=273+15= 288 K

V1 =34.2 ml

P2 =760 mm

V2 =? ml

T2 =273K

P1 V1 /T1 =P2 V2 /T2

(737 x 34.2) /288 =(760 x V2) /273

V= 31.44 ml.

Vapour Density =wt.of  vapours or substance /(volume of vapours at  N.T.P. X 0.00009 )

= 0.17 /(31.44 x 0.00009)

= 60.07

Mol. wt. = 2×V.D. = 2×60=120 Ans.

Q—-0.1133 gm. org. compd on heating displaces 22.8 ml air at 150C & pre. is 750 mm (aq. tension at 15 0C=13 mm) Calculate the Mol. wt.

Ans.

P1 = 750-13=737 mm

T1=273+15= 288 K

V1 =22.8 ml

P2 =760 mm

V2 =? ml

T2 =273K

P1 V1 /T1 =P2 V2 /T2

(737 x 22.8) /288 =(760 x V2) /273

V= 20.96  ml.

Vapour Density =wt.of  vapours or substance /(volume of vapours at  N.T.P. X 0.00009 )

= 0.1133 /(20.96 x 0.00009)

=60

Mol. wt. = 2×V.D. = 2×60=120 Ans.

Elevation in Boiling point-

M =(1000 Kb.w) / ΔT.W

Kb =molal elevation constt.

w   =wt. of solute  in gm.

W   =wt. of solvent in gm.

ΔT    = Elevation in b.p.

Also

M =(100 Kb.w) / ΔT.W

Kb= Molar or Molecular elevation constt.

Depression in Freezing Point

M =(1000 Kf.w) / ΔT.W

Kf= molal depression constt.

Δ T= depression in freezing pt.

Also

M =(100 Kf.w) / ΔT.W

=Molar depression constt.