Calculate the mass of sodium acetate \((\text{CH}_3\text{COONa})\) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is \(82.0245\, \text{g mol}^{–1}.\)
Solution and Explanation
0.375 M aqueous solution of sodium acetate
= 1000 mL of solution containing 0.375 moles of sodium acetate
∴Number of moles of sodium acetate in 500 mL
\(= \frac{0.375 }{ 1000} × 500\)
= 0.1875 mole
Molar mass of sodium acetate = \(82.0245\, \text{g mole}^{-1}\) (Given)
∴ Required mass of sodium acetate \(=82.0245\, \text{g mole}^{-1}\) \((0.1875\ \text{mole})\)
= 15.38 g
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Concepts Used:
Molecular Mass of Polymers
It is described as the distribution rather than a specific number due to the occurrence of polymerization in such a way as to produce different chain lengths. Polymer MW is derived as follows:
\[M_{W} = \sum^{N}_{i=1} w_{i}MW_{i}.\]Where,
wi = the weight fraction of polymer chains having a molecular weight of MWi.
The MW is typically measured by light dispersing experiments. The degree of dispersing arises from the molecule size and, thus, molecular weight dispensation can be mathematically set on the total scattering created by the sample.