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In a test-tube, there is 18 g of glucose `(C_(6)H_(12)O_(6))` 0.08 mole of glucose is taken out. Glucose left in the test tube isA. 0.10 gB. 0.02 gC. 0.10 molD. 3.60 g |
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Answer» Correct Answer - D Molar mass of glucose, `C_(6)H_(12)O_(6) = 180 g mol^(-1)` 18 g glucose = 0.10 mol Moles of glucose left = 0.10-0.08 = 0.02 mol Mass of glucose = `(0.02 mol)xx(180 g mol^(-1)) = 360g` |
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