(i) \(\text{H}_2\text{O}\):
The molecular mass of water, \(\text{H}_2\text{O}\)
\(= (2 \times \text{Atomic mass of hydrogen}) + (1 \times \text{Atomic mass of oxygen})\)
= [2(1.0084) + 1(16.00 u)]
= 18.016
= 18.02 u
(ii) \(\text{CO}_2\):
The molecular mass of carbon dioxide, \(\text{CO}_2\)
\(= (1 \times \text{Atomic mass of carbon}) + (2 \times \text{Atomic mass of oxygen})\)
= [1(12.011 u) + 2 (16.00 u)]
= 12.011 u + 32.00 u
= 44.01 u
(iii) \(\text{CH}_4\):
The molecular mass of methane, \(\text{CH}_4\)
\(= (1 \times \text{Atomic mass of carbon}) + (4 \times \text{Atomic mass of hydrogen})\)
= [1(12.011 u) + 4 (1.008 u)]
= 12.011 u + 4.032 u
= 16.043 u
Calculate the number of moles present in 9.10 × 1016 kg of water.
0.1 mole of compound S will weigh ...... g, (given the molar mass in g mol\(^{-1}\) C = 12, H = 1, O = 16)
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.