Answer:
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Two mixtures were prepared from three very narrow molar mass distribution polystyrene samples with molar masses of 10,000, 30,000 and 100,000 g mol−1 as indicated below: (a) Equal numbers of molecules of each sample (b) Equal masses of each sample (c) By mixing in the mass ratio 0.145:0.855 the two samples with molar masses of 10,000 and 100,000 g mol−1 For each of the mixtures, calculate the number-average and weight-average molar masses and comment upon the meaning of the values.
Answers:
a) 46.666.66 g/mol
b) 20930.23 g/mol
c)43333.33 g/mol
Explanation:
A)The equal number of molecules of each sample can be calculated using [tex]Mn = \frac{n(M1 + M2 + M3)}{3n}[/tex]
because for the number of molecules to be equal : n1 = n2 = n3 = n
Mn = 46666.66 g/mol
B ) To calculate the equal masses of each sample
we apply this equation
[tex]Mn = \frac{W1 + W2 +W3}{\frac{W1}{M1} +\frac{W2}{M2}+ \frac{W3}{M3} }[/tex]
ATTACHED BELOW IS THE REMAINING PART OF THE SOLUTION
