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18 g of glucose is dissolved in 90 g of water. The relative lowering of vapour pressure is equal to ….

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Answer: 0.1

\(\frac{P°-P}{P°}\) = x2

x= No. of moles of glucose

\(\frac{18}{180}\) = 0.1

\(\frac{P°-P}{P°}\) = 0.1

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