Assuming total absorption of the light by the sample, what is the maximum amount (in moles) of CH3X that breaks apart when a cuvette containing a solution of CH3X is irradiated with 280-nm light with a power of 885 mW for 10.0 minutes, given that the quantum yield for the reaction CH3X → CH3 + X is f = 0.24?

Verified step by step guidance

1. Calculate the energy of a single photon using the formula: \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant (6.626 \times 10^{-34} \text{ J s}), \( c \) is the speed of light (3.00 \times 10^8 \text{ m/s}), and \( \lambda \) is the wavelength (280 \text{ nm} = 280 \times 10^{-9} \text{ m}).
2. Determine the total energy supplied by the light source using the formula: \( \text{Total Energy} = \text{Power} \times \text{Time} \), where the power is 885 mW (or 885 \times 10^{-3} \text{ W}) and the time is 10.0 minutes (or 600 seconds).
3. Calculate the total number of photons using the formula: \( \text{Number of Photons} = \frac{\text{Total Energy}}{E} \), where \( E \) is the energy of a single photon calculated in step 1.
4. Use the quantum yield (\( f = 0.24 \)) to find the number of moles of CH3X that dissociate. The formula is: \( \text{Moles of CH3X} = \frac{\text{Number of Photons} \times f}{N_A} \), where \( N_A \) is Avogadro's number (6.022 \times 10^{23} \text{ mol}^{-1}).
5. The result from step 4 gives the maximum amount of CH3X in moles that breaks apart under the given conditions.