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# Physical chemistry–A-1

 Question: – calculate the total n of atoms in 78gm of k Solution: – 6.022×1023 atom =39gm of k 78gm of k = 6.022×1023 atom k=12.044×1023 atom Question: – Calculate the total no of decimoles in 46gm of Na Solution: – 1 mole of Na = 23gm 46gm = 2mole 1mole = 10decimole 2mole =20decimoele Use of mole concept of compound· Particle = molecule Formula wt. =mol wt. Name of mole =gm mole or, gm molecular One gm mole CaCo3 = one mole CaCo3 6.022×1023 /H1H2 molecular = 100gm One gm –atom of hydrogen = 1gm mole one gm –mole of hydrogen = 2gm mole one mole Na2Co3 = one gm mole of Na2Co3 =6.022×1023 molecular =106gm total no of e- ion one molecular of NaCo3 =6atom one molecular of NaCo3 =2atomof Na = 1atom of Ca = 3atom of o no of e- in one molecular of NaCo3 = 11×2+6+8×3 =52e- Question: – calculate the total no of atom in 49gm f H2So4 Solution: – one mole H2So4 = 6.022x1023molecular =98gm 49gm = 6.022×1023/2 molecular = 3.022x1023molecular Total no atom of in H2So4 = 7atom Question: -calculate the total no of e- in 50gm caco3 Solution: – one mole caco3 = 6.022x1023molecular =40+12+16×3=100gm 50gm = 6.022x1023x56/100= 3.011x1023moecular Total no e- in ca = 20 Page no.- 04 C = 6 O = 24 50e- Total no of e-in = 3.022x1023x50e- Use of me concept for radicle· Particle = ion Formula wt. = ionic wt. Moe = gm – ion One mole Na+=one gm –ion Na+ = 6.022x1023ion = 23gm Na Na+ 11e- 10e- 11p 11p 12N 12p A = 11p+12p A = 11p+12n No of electron in one ion of Na+ =10e- Question – calculate the total no of e-in 48gm of So4. – Solution – one mole of So4 = 32+16×4 = 96gm =6.022×1023 =48gm = 3.022x1023ion No of e- one ion of So4 =16+8×4+2 = 50e- Mole in terms of volume (only for gaseous substance) · Molar volume: – volume of one mole nature gas at any temperature and pressure is called molar volume. Case1: – Molar volume of S.T.P or N.T.P T =00C or 273k p = 1atom Molar volume at S.T.P = 22.4 lit =22400cc(or) ml One gm-mole oxygen at S.T.P = 22.4 lit
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