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 $C_6{H}_{12}{O}_6$. 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 $1gc{m}^{-3}$.
Rreaction that takes place during photosynthesis is:
$6C{O}_2+6H_{2}O$ $⟶$ $C_6{H}_{12}{O}_6+6O_2$

Answer 1

Molecular mass of $6H_{2}O$ = 6[(1 x 2) + 16)] =6 x 18 =108g
Molecular mass of $C_6{H}_{12}{O}_6$ = 12 x 6 + 1 x 12 +16 x 6 = 180g

180 g of glucose needs $(6\times 18)g$ of water
Therefore, 18 g of glucose will need $\frac{108}{180}\times 18g$ of water = 10.8 g

Given, density of water to be $1gc{m}^{-3}$
mass of water = 10.8 g
Therefore, Volume of water used $=\frac{Mass}{Density}=\frac{10.8g}{1gc{m}^{-3}}=10.8c{m}^3$