Question

In photosynthesis, 6 molecules of carbon dioxide combine with an equal number of water molecules through a complex series of reactions to give a molecule of glucose having a molecular formula ${\mathrm{C}}_6{\mathrm{H}}_{12}{{\mathrm{O}}_6}_{}$ and a molecule of oxygen with a molecular formula ${\mathrm{O}}_2$ . How many grams of water would be required to produce 18 g of glucose? Compute the volume of water so consumed assuming the density of water to be 1 g $c{m}^{-3}$.

Answer 1

We know that

The reaction of photosynthesis :

$6{\mathrm{CO}}_2\left(\mathrm{g}\right)+6{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to {\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\left(\mathrm{s}\right)+6{\mathrm{O}}_2\left(\mathrm{g}\right)$

• 1 mole of Glucose needs 6 moles of water
• 180 g of Glucose needs ($6\times 18)$ of water.
• 1 g of glucose will need $\frac{108}{180}g$ of water
• 18 g of Glucose would need $\frac{108}{180}\times 18g$of water=$10.8g$
• Volume of water used =$\frac{Mass}{density}=\frac{10.8g}{1g}c{m}^{-3}=10.8c{m}^3$