Education

Physical chemistry–A-3.

No of mole = no of particles given / 6.022×1023
Relative density·
(i) Vapor density (V.D) (for gaseous substance)
V.D =density of gaseous substance/density of hydrogen
= wt. of one lit gaseous substance x22.4S.T.P /of one lit hydrogen
x22.4 S.T. P
m.wof gaseous substance /M. W of hydrogen
V.D =M/2
(2) Specific gravity (for liquid and solid)
Density of a substance
Density of water
density of water =1gm/cc
SP gravity = density of the substance
SP gravity = no unit
SP gravity = 1.12
Density =1.12gm /cc
N0TE
Specific volume: -·
Volume per unit mass is called specific volume.
Specific volume = ½
Unit = m3/kg, it /gm, cc/gm
Equivalent wt. of element: -·
That wt. of element which combine with or displace form.
– 1 part by wt. of hydrogen
M.W =2XV.D
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– 8 part by wt. of oxygen
– 35.5 part by wt. of chlorine
Equivalent wt. by hydrogen·
M + H MH (metal hydrogen)
X gm y gm z gm = x + y
Y gm hydrogen = x gm metal
1 gm hydrogen = x/y gm metal
ONE gm mole hydrogen = 2 gm
2 gm- atom = NA molecular = 2 NA atom
= 22.4 lit S.T.P = 24.45 lit R.T.P
PV = W/M RT
Equivalent wt. by oxygen·
M + O MO (metal oxide)
X gm y gm z gm = x+ y
Y gm oxygen = x gm metal
8 gm oxygen = x/y x8 metal
One gm mole oxygen = 32 gm = 2 gm-atom
= NA molecular = 2NA atom = 22.4 lit
S.T.P =24.45 lit R.T.P
E.M = X/Y
E.M =X/Y X8

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PV = W/M RT
Equivalent wt. by chlorine: -·
M + CL MCL (metal chlorine)
X gm y gm z gm = x + y
Y gm chlorine = x gm metal
35.5 gm chlorine = x / y x 35.5 metal
One gm mole chlorine = 71 gm
= 2 gm-atom = NA molecular
= 2 NA atom = 22.4 lit S.T.P = 22.45lit R.T.P
PV = W/M RT
Note
M + H2SO4 MSO4 + H2
X gm y gm
Question: – 3.2 gm of metal reacts with oxygen to give 4 gm metal oxide
calculate equation wt. f metal.
M + O MO
3.2 gm 0.8 gm 4 gm
E.M = X/Y X8 , 3.2 / 0.8 X 8 = 32 gm
E.M = X/Y X35.5
E.M = X/Y

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