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Section: Chemistry

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Question : 4 of 29

Marks: +1, -0

In aqueous glucose solution, the mole fraction of water is 40 times to mole fraction of glucose. What is the weight percentage

\left(\frac{w}{w}\right)

of glucose in the solution ?

[AP EAPCET 23 May 2025 S2]

$40$
30
20
10

Solution:

Mole fraction of binary solution,

\chi_{\text{glucose }} + \chi_{\text{water }} = 1

\chi_{\text{glucose }} + 40 \times \chi_{\text{glucose }} = 1

\chi_{\text{glucose }} = \frac{1}{41}

Assume 1 mole of total solution Moles of glucose

(n_{\text{glucose }}) = \chi_{\text{glucose }} \times 1 = \frac{1}{41} \text{ mol}

Moles of water

(n_{\text{water }}) = \frac{40}{41} \text{ mol}

Mass of glucose (

M_{\text{glucose }}

)

= n_{\text{glucose }} \times \text{ Molar mass }

= \frac{1}{41} \times 180 = \frac{180}{41}

Mass of water

(m_{\text{water }}) = n_{\text{water }} \times

Molar mass

= \frac{40}{41} \times 18 = \frac{720}{41}

\text{w/w \%}

of glucose

\frac{\text{ Mass of glucose }}{\text{ Total mass }} \times 100

= \text{ Total mass } = \frac{180}{41} + \frac{720}{41} = \frac{900}{41} \text{ g}

Weight \\% of glucose

= \frac{\frac{180}{41}}{\frac{900}{41}} \times 100 \% = 20 \%

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