Homework Answers
a) Molar mass of ( NH4)2S2O8 = 228 g/ mol.
Molarity = W2*1000/M2*V. (1)
Where, W2 = mass of solute[ ( NH4)2S2O8]
M2 = molar of solute = 228 g/mol
V = Volume of solution in mL = 125 mL
Now , putting the above value in the Eq. 1
1.5 = W2*1000/228*125
Or, W2 = 1.5*228*125/1000 = 42.75 g.
Hence, 42.75 g of (NH4)2S2O8 is required.
b) Let final Concentration of the resulting solution after mixing = S3
Applying , the equation
V1S1 + V2S2 = V3S3
63.54*0.500 + 125*1.5 = (125+63.54)*S3
Or, S3 = (31.77 + 187.5)/188.54 = 1.16 M
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