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CONCEPT of important formulac of physical chemistry. Some Basic Concepts of Chemistry Structure of Atom Actual yield © Radius of nucleus (R) = RyA'” Percent ~ TR aEaMRERES eserntinl TE 00 > CE ee cake ed Ro = Constant (= 1.33 107” cm) Mass of one atom of the clement A= Mass number of atom ° Ani 885° ———————— rae he ' 75 * Mass of an atom of C~12 . De ps iE = —shvehvy+ mv = Eq. mass x Valency u , : 7 ° 5 =109.677( 5-5) cm"; =3.29x10!( 4-4) e Alomic mass = Molecularmass 64 nm ny nm ~ Atomicity Specific heat (cal/g) 52.98 @ Molecular mass = 2 x Vapour density . Horse oad rs = M9 oe ——— pm ® Number of gram atoms = Gram atomic mass (GAM) . E,= “Rul Z) = -2isx10"( 2) J/atom w (g) nn” @ Number of gram molecules = molecular \ 2 _7 vane io ia or ~4- «1312 ki/mol or —= —:E=—=hv=hvy + mv = Eq. mass x Valency : Hy Molecular mass 64 e@ v= 109,677 =~] em“';y=329x1085( 1-4) “Ss e Atomic mass = ——— = ——_______—__ m My mn; ony Atomicity Specific heat (cal/g) i 2 9 @ Molecular mass = 2 x Vapour density © mvr= ra es or 7 pm 5 e Number of gram atoms = —— Via Zz Gram atomic mass (GAM) ° E,= -R,{ 2] =-2.18x 102) J/atom w (g) n n @ Number of gram molecules = —— G lecular mass (GMM -z' =" eta a ( ) or as 1312 kJ/mol or 2 x 13.6 eV/atom © Number of gram equivalents = ————_—_ ed id Gram equivalent mass (GEM) ‘ 2.19x10°xZ 5 : Atomic mass °v.=—""""" ms @ Equivalent mass of an element =—————— n Valency > Molecular mass of acid ste a on -is( | 1 Equivalent f a e AE = kyl 2 + |=218%10 = @ Equivalent mass of an aci Basicity non re rm e ER Ta SO ection e toe h “< h Acidity my p 2mE, 2mqV e Percentage composition = pane Cis eet x 100 h h Become Molar mass of compound e ae Re a or Ax- Ave — Mass of solute ta ; * Mass%= — x 100 © Schrédinger wave equation: Mass of solution : r 2 8x°m z © Molarity(M) = Bx — Vy +—=—(E-V)y = 0; V = Laplacian operator Mz x V (in mL) h We x 1000 @ Maximum no. of spectral lines produced when an electror eee coe A = @ Normality (N) = EM, x V Gin mL) es n(n—1) ® Molality(m)= ee @ No. of lines in the spectrum when an electron returns from Mz, x W, (ing) sn, amr =m +0 where, W, = Mass of solute, Mp = Molecular mass of solute ce AR fem > W, = Mass of solvent a2 npg Na e L.E.= E.. — FE, =2.18x10 x—> J/atom @ Mole fraction, xp = and x, = n nat np nya + np 2 M, V\/ny = M,V,/n, orN) Vv; = N2V> M,( V; + V2) =M, Vv; +M3V, 1 M 1000d 1000 =|—-—2y = Xp M, M,xM mxM, Number of moles = Molarity x Volume Number of equivalents = Normality x Volume Molecular formula =n x Empirical formula Molecular formula mass r Empirical formula mass n or 13.6x ate eV/atom a? ® Orbital angular momentum = ,/(/ + = x @ Magnetic moment = Vn(n +2) B.M. e Spin angular momentum = ,/s(s + p= 2n @ Spin multiplicity = 2S + 1; $ = Sum of spin quantum numbers. @ Radial nodes =n -!-1; Angular nodes =! e Total no. ofnodes=n-1 
