c) The recommended daily intake of vitamin \( \mathrm{D} 3\left(\mathrm{C}_{27} \mathrm{H}_{44} \mathrm{O}\right) \) is about 20 i) Calculate the moles of Vit \( \mathrm{D}_{3} \) in \( 20 \mu \mathrm{g} \mathrm{C}_{27} \mathrm{H}_{44} \mathrm{O} \).

To calculate the moles of vitamin \( \mathrm{D}_{3} \) in \( 20 \mu \mathrm{g} \mathrm{C}_{27} \mathrm{H}_{44} \mathrm{O} \), we need to use the molar mass of vitamin \( \mathrm{D}_{3} \) and the given mass of \( \mathrm{C}_{27} \mathrm{H}_{44} \mathrm{O} \). Given: - Molar mass of vitamin \( \mathrm{D}_{3} \) (\( \mathrm{C}_{27} \mathrm{H}_{44} \mathrm{O} \)): 396.6 g/mol - Mass of \( \mathrm{C}_{27} \mathrm{H}_{44} \mathrm{O} \): 20 μg First, we need to convert the mass of \( \mathrm{C}_{27} \mathrm{H}_{44} \mathrm{O} \) from micrograms to grams: \[ 20 \mu \mathrm{g} = 20 \times 10^{-6} \mathrm{g} \] Now, we can calculate the moles of vitamin \( \mathrm{D}_{3} \) using the formula: \[ \text{Moles} = \frac{\text{Mass}}{\text{Molar mass}} \] Substitute the values: \[ \text{Moles} = \frac{20 \times 10^{-6} \mathrm{g}}{396.6 \mathrm{g/mol}} \] Let's calculate the moles of vitamin \( \mathrm{D}_{3} \) in \( 20 \mu \mathrm{g} \mathrm{C}_{27} \mathrm{H}_{44} \mathrm{O} \). Calculate the value by following steps: - step0: Calculate: \(\frac{20\times 10^{-6}}{396.6}\) - step1: Convert the expressions: \(\frac{20\times 10^{-6}}{\frac{1983}{5}}\) - step2: Evaluate the power: \(\frac{\frac{1}{50000}}{\frac{1983}{5}}\) - step3: Multiply by the reciprocal: \(\frac{1}{50000}\times \frac{5}{1983}\) - step4: Reduce the numbers: \(\frac{1}{10000}\times \frac{1}{1983}\) - step5: Multiply the fractions: \(\frac{1}{10000\times 1983}\) - step6: Multiply: \(\frac{1}{19830000}\) The moles of vitamin \( \mathrm{D}_{3} \) in \( 20 \mu \mathrm{g} \mathrm{C}_{27} \mathrm{H}_{44} \mathrm{O} \) is approximately \( 5.042864 \times 10^{-8} \) moles.