1.

How much bacl2 would be needed tomake 250ml of a solution having same concentration of cl- ions as one containing 3.78 gm of nacl per 100ml of the solution?

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-]=w*1000/208.3*250

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

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



Discussion

No Comment Found