The mole fractions of glucose (C_6H_{12}O_6) | vedclass

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(C) The mole fraction of a component is given by $x_i = \frac{n_i}{n_{total}}$.Let the total number of moles be $n_{total} = 1 \ mol$.Then,the moles o

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$20$

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(C) The mole fraction of a component is given by $x_i = \frac{n_i}{n_{total}}$.Let the total number of moles be $n_{total} = 1 \ mol$.Then,the moles of glucose $(n_{glucose})$ = $0.0244 \ mol$ and the moles of water $(n_{water})$ = $0.9756 \ mol$.The molar mass of glucose $(C_6H_{12}O_6)$ is $180 \ g/mol$ and the molar mass of water $(H_2O)$ is $18 \ g/mol$.Mass of glucose = $0.0244 \ mol 	imes 180 \ g/mol = 4.392 \ g$.Mass of water = $0.9756 \ mol 	imes 18 \ g/mol = 17.5608 \ g$.Total mass of the solution = $4.392 \ g + 17.5608 \ g = 21.9528 \ g$.Weight percentage $(w/w)$ of glucose = $\frac{	ext{mass of glucose}}{	ext{total mass of solution}} 	imes 100 = \frac{4.392}{21.9528} 	imes 100 \approx 20\%$.

The mole fractions of glucose $(C_6H_{12}O_6)$ and water $(H_2O)$ in an aqueous glucose solution are $0.0244$ and $0.9756$ respectively. What is the weight percentage $(w/w)$ of glucose in this solution?

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