Chapter 27: Problem 4

Number average molecular mass is given by the formula (a) \(\bar{M}_{n}=\frac{\Sigma M_{f}}{\Sigma N_{A}}\) (b) \(\bar{M}_{n}=\frac{\sum N_{1} M_{f}}{\Sigma N_{l}}\) (c) \(\bar{M}_{n}=\frac{\sum N_{i} M_{f}^{2}}{\sum M_{l}}\) (d) \(\bar{M}_{n}=\frac{\sum N_{1} M_{f}^{2}}{\Sigma M_{l}}\)

Short Answer

Expert verified

None of the given options provide the correct formula for the number average molecular mass.

Step by step solution

Comparing Option (a)

Looking at formula a, the correct notation for the total mass of all molecules, \(\Sigma M_{f}\), is in the numerator. However, the denominator should be the total number of molecules not Avogadro's number, \(\Sigma N_{A}\). Hence option (a) is incorrect.

Comparing Option (b)

For formula b, the sum of all mole fractions multiplied by their respective molecular masses, \(\sum N_{1} M_{f}\), is in the numerator and the denominator is the sum of mole fractions \(\Sigma N_{l}\). This describes the weight average molecular weight, not the number average molecular weight. Hence, option (b) is incorrect.

Comparing Option (c)

Looking at formula c, the numerator \(\sum N_{i} M_{f}^{2}\) implies squaring of molecular mass which is not done in number average molecular mass. Hence this formula is not correct.

Comparing Option (d)

Considering formula d, similar to (c), squaring of molecular mass is not acceptable. Hence, option (d) is incorrect.

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Number average molecular mass is given by the formula (a) \(\bar{M}_{n}=\frac{\Sigma M_{f}}{\Sigma N_{A}}\) (b) \(\bar{M}_{n}=\frac{\sum N_{1} M_{f}}{\Sigma N_{l}}\) (c) \(\bar{M}_{n}=\frac{\sum N_{i} M_{f}^{2}}{\sum M_{l}}\) (d) \(\bar{M}_{n}=\frac{\sum N_{1} M_{f}^{2}}{\Sigma M_{l}}\)

Short Answer

Step by step solution

Comparing Option (a)

Comparing Option (b)

Comparing Option (c)

Comparing Option (d)

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