(1) Given, Mass concentration of A=2.59g/dm^-3 Mass of concentration B=mass/volume in dm^3 =3.01/0.25 =12.04g/dm^-3 volume of base(VB)=25.0cm^3 volume of acid:(VA)=30.50cm^3 (1ai) Molar mass of A(HCl) =1+35.5=36.5g/mol^1 Concentration in mol/dm3=concentration in g/dm3^1 /molar mass =2.52/36.5 CA=0.069mol/dm^3 (1aii) Concentration of B in g/dm^3 =mass in g/volume in dm^-3 =3.01/0.25 =12.04g/dm^3 (1aiii) Mole of ratios of acid and base nA=2, nB=1 Using; CAVA/CBVB=nA/nB (0.069*30.50)/(CB*25.0)=2/1 CB:(0.069*30.50*1)/25.0*2 CB=2.1045/50 =0.042mol/dm^3 Concentration of B in mol/dm^3 is 0.042mol/dm^3 (1bi) Molar mass of Na2Co3 . xH2O = (23*2)+(12)+(16*3)+ x(1*2+16) =46+12+48+18x =106+18x 0.042=12.04/x x=12.04/0.042=286.7gmol^-1 286.7=106+18x 18x=286.7 -106 18x=180.7 x=180.7/18 x=10 The number of molecules of water of crystallization present is 10. (1bii) The formular is Na2Co3 .xH2O= Na2CO3 . 10H2O
Posted on: Wed, 17 Sep 2014 10:06:04 +0000
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