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# 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 Page no.- 08 – 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 Page no.- 09 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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