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Solutions question

2023 · 15 Apr · Shift 1 · Q19
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Solutions question

2023 · 15 Apr · Shift 1 · Q19

JEE MainChemistrySolutionsNumerical+4 / −1
The vapour pressure of 30%(w/v)30 \%(\mathrm{w} / \mathrm{v})30%(w/v) aqueous solution of glucose is ‾\underline{\hspace{2cm}}​mm Hg\mathrm{mm} ~\mathrm{Hg}mm Hg at 25∘C25^{\circ} \mathrm{C}25∘C. [Given : The density of 30%30 \%30%(w/v), aqueous solution of glucose is 1.2 g cm−31.2 \mathrm{~g} \mathrm{~cm}^{-3}1.2 g cm−3 and vapour pressure of pure water is 24 mm Hg24 \mathrm{~mm}~ \mathrm{Hg}24 mm Hg.] (Molar mass of glucose is 180 g mol−1180 \mathrm{~g} \mathrm{~mol}^{-1}180 g mol−1.)
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