How much bacl2 would be needed to make 250 ml of a solution having the same concentration of cl minus as one containing?
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Let's take the case of NaCl.
NaCl->Na+ + Cl-
[Cl-]=[NaCl]=2.92*1000/58.5*100=0.5M
{Molarity=weight(in g)*1000/Mol.mass(in g/mol)*volume}
Now, BaCl2.
BaCl2->Ba2+ + 2Cl-
[Cl-]=0.5M
[BaCl2]=[Cl-]/2=0.25M
0.25=w*1000/208.3*250
{Mol. mass of BaCl2=208.3, Ba=137.3, Cl=35.5}
w=0.25*208.3*250/1000=13.02g
Molecules dissociated as I have assumed the solvent is water or any polar solvent.
NaCl->Na+ + Cl-
[Cl-]=[NaCl]=2.92*1000/58.5*100=0.5M
{Molarity=weight(in g)*1000/Mol.mass(in g/mol)*volume}
Now, BaCl2.
BaCl2->Ba2+ + 2Cl-
[Cl-]=0.5M
[BaCl2]=[Cl-]/2=0.25M
0.25=w*1000/208.3*250
{Mol. mass of BaCl2=208.3, Ba=137.3, Cl=35.5}
w=0.25*208.3*250/1000=13.02g
Molecules dissociated as I have assumed the solvent is water or any polar solvent.
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