A) 110000 and 10538
B) 36000 and 43333
C) 140000 and 156000
D) 228000 and 56000
Correct Answer: B
Solution :
\[{{\overline{M}}_{n}}\] (number average molecular mass) \[=\frac{30\times 20000+40\times 30000+30\times 60000}{(30+40+30)}\]\[=36000\] \[{{\overline{M}}_{w}}\] (weight average molecular mass) \[30\times {{(20000)}^{2}}+40\times {{(30000)}^{2}}\] \[=\frac{+30\times {{(60000)}^{2}}}{30\times 20000+40\times 30000+30\times 60000}\]\[=43333\]You need to login to perform this action.
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