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(D) The balanced chemical equation for photosynthesis is: $6CO_2 + 6H_2O \xrightarrow{	ext{Chlorophyll, Sunlight}} C_6H_{12}O_6 + 6O_2$.From the equa

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$10.8$

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(D) The balanced chemical equation for photosynthesis is: $6CO_2 + 6H_2O \xrightarrow{	ext{Chlorophyll, Sunlight}} C_6H_{12}O_6 + 6O_2$.From the equation,$1$ mole of glucose $(C_6H_{12}O_6)$ is produced from $6$ moles of water $(H_2O)$.The molar mass of glucose is $(6 	imes 12) + (12 	imes 1) + (6 	imes 16) = 72 + 12 + 96 = 180\, g/mol$.The molar mass of water is $(2 	imes 1) + 16 = 18\, g/mol$.Thus,$180\, g$ of glucose requires $6 	imes 18 = 108\, g$ of water.Therefore,$18\, g$ of glucose requires: $\frac{108}{180} 	imes 18 = 10.8\, g$ of water.Given the density of water is $1\, g\, cm^{-3}$,the volume of water consumed is: $	ext{Volume} = \frac{	ext{Mass}}{	ext{Density}} = \frac{10.8\, g}{1\, g\, cm^{-3}} = 10.8\, cm^3$.

Species property	$H_2O$	$CO_2$	$Na$ atom	$MgCl_2$
No. of moles	$2$	-	-	$0.5$
No. of particles	-	$3.011 \times 10^{23}$	-	-
Mass	$36 \, g$	-	$115 \, g$	-

Similar Questions

Fill in the missing data in the Table.
Species property $H_2O$ $CO_2$ $Na$ atom $MgCl_2$
No. of moles $2$ - - $0.5$
No. of particles - $3.011 \times 10^{23}$ - -
Mass $36 \, g$ - $115 \, g$ -

Cinnabar $(HgS)$ is a prominent ore of mercury. How many grams of mercury are present in $225 \,g$ of pure $HgS$ (in $,g$)? Molar mass of $Hg$ and $S$ are $200.6 \,g \,mol^{-1}$ and $32 \,g \,mol^{-1}$ respectively.

Which of the following species have $18$ electrons? $Ca^{2+}, K^{+}, Na, Cl^{-}, Ar$. Also,state the reason why the chemical properties of all isotopes of an element are similar.

Calculate the number of molecules in $0.5$ mole of water.

Calculate the number of particles in each of the following:
$(a)$ $48 \,g$ of $Mg$
$(b)$ $8 \,g$ of $O_2$
$(c)$ $0.1$ mole of carbon (Atomic mass $Mg = 24 \,u, O = 16 \,u, C = 12 \,u$,Avogadro constant $= 6.022 \times 10^{23}$)

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