1.

how much bacl2 would be needed to make 250 ml of a solution having same concentration of cl- as the one containing 3.78g of nacl per 100 ml?

Answer»

Let's take the CASE of NaCl.

NaCl->Na+ + Cl-

[Cl-]=[NaCl]=3.78*1000/58.5*100=?

{Molarity=weight(in g)*1000/Mol.mass(in g/mol)*volume}

Now, BaCl2.

BaCl2->BA2+ + 2CL-

[Cl-]=?

[BaCl2]=[Cl-]/2=?

[Cl-]/2=w*1000/208.3*250

{Mol. mass of BaCl2=208.3, Ba=137.3, Cl=35.5}

w=[BaCl2]*208.3*250/1000=?


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