In an aqueous glucose solution,the mole fract | vedclass

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(C) Let $n_g$ be the moles of glucose and $n_w$ be the moles of water. Given that the mole fraction of water $(x_w)$ is $40$ times the mole fraction o

Vedclass · Accepted Answer

$20$

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(C) Let $n_g$ be the moles of glucose and $n_w$ be the moles of water. Given that the mole fraction of water $(x_w)$ is $40$ times the mole fraction of glucose $(x_g)$,we have $x_w = 40 x_g$. Since $x_w + x_g = 1$,we get $40 x_g + x_g = 1$,which implies $41 x_g = 1$,so $x_g = 1/41$ and $x_w = 40/41$. The ratio of moles is $n_w / n_g = x_w / x_g = 40/1$. Thus,$n_w = 40 n_g$. The mass of glucose $(m_g)$ is $n_g 	imes 180 \ g/mol$ and the mass of water $(m_w)$ is $n_w 	imes 18 \ g/mol = 40 n_g 	imes 18 \ g/mol = 720 n_g \ g$. The weight percentage of glucose is $(m_g / (m_g + m_w)) 	imes 100 = (180 n_g / (180 n_g + 720 n_g)) 	imes 100 = (180 / 900) 	imes 100 = (1/5) 	imes 100 = 20\%$. Therefore,the correct option is $C$.

In an aqueous glucose solution,the mole fraction of water is $40$ times the mole fraction of glucose. What is the weight percentage $(w/w)$ of glucose in the solution?

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