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18 g of glucose is dissolved in 90 g of water. The relative lowering of vapour pressure is equal to …. |
Answer» Answer: 0.1 \(\frac{P°-P}{P°}\) = x2 x2 = No. of moles of glucose \(\frac{18}{180}\) = 0.1 \(\frac{P°-P}{P°}\) = 0.1 |
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