Percentage of Carbon:
The mass of carbon in 0.2 g of CO$_2$ can be calculated using the molar masses: Molar mass of CO$_2$ = 44 g/mol Molar mass of C = 12 g/mol
Mass of C in 0.2 g CO$_2 = \frac{12}{44} \times 0.2 g = 0.0545g$
Percentage of C $= \frac{Mass of C}{Mass of compound} \times 100 = \frac{0.0545}{0.3} \times 100 = 18.18%$
Percentage of Hydrogen:
The mass of hydrogen in 0.1 g of H$_2$O can be calculated using the molar masses: Molar mass of H$_2$O = 18 g/mol Molar mass of H = 1 g/mol (but there are 2 H atoms, thus, 2 g/mol)
Mass of H in 0.1 g H$_2$O $= \frac{2}{18} \times 0.1g = 0.0111g$
Percentage of H $= \frac{\text{Mass of H}}{\text{Mass of compound}} \times 100 = \frac{0.0111}{0.3} \times 100 = 3.70%$
Therefore, the percentage composition of carbon and hydrogen is 18.18% and 3.70%, respectively.
What is the correct IUPAC name of the following compound?
Choose the correct option for structures of A and B, respectively:
Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): A typical unfertilized, angiosperm embryo sac at maturity is 8-nucleate and 7-celled.
Reason (R): The egg apparatus has 2 polar nuclei.
In the light of the above statements, choose the correct answer from the options given below: