Two mixtures were prepared from three very narrow molar mass distribution polystyrene samples with molar masses of 10,000, 30,00

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Two mixtures were prepared from three very narrow molar mass distribution polystyrene samples with molar masses of 10,000, 30,00

0 and 100,000 g/mola indicated below: (A) Equal numbers of molecules of each sample (B) Equal masses of each sample. For each of the mixtures, calculate the number-average and weight-average molar masses and comment upon the meaning of the values.

Chemistry

1 answer:

alisha [2.9K] · Answer 1 · 2026-03-10T20:53:20+03:00

Answer:

Some components of your question seem to be missing.

We created two mixtures from three polystyrene samples with very narrow molar mass distributions, which are 10,000, 30,000, and 100,000 g mol−1, as outlined below: (a) Equal molecule counts of each sample (b) Equal sample masses (c) Mixing the two samples with molar masses of 10,000 and 100,000 g mol−1 in a mass ratio of 0.145:0.855. For each mixture, compute the number-average and weight-average molar masses and interpret the significance of these values.

Answers:

a) 46,666.66 g/mol

b) 20,930.23 g/mol

c) 43,333.33 g/mol

Explanation:

A) For equal numbers of molecules from each sample, we use $Mn = \frac{n(M1 + M2 + M3)}{3n}$

to determine equal molecule counts: n1 = n2 = n3 = n.

The resulting Mn = 46,666.66 g/mol.

B) To ascertain equal masses of each sample, we apply the following formula:

$Mn = \frac{W1 + W2 +W3}{\frac{W1}{M1} +\frac{W2}{M2}+ \frac{W3}{M3} }$

THE REMAINING PORTION OF THE SOLUTION IS ATTACHED BELOW