1.

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 is

Answer»

0.10 G
0.02 g
0.10 mol
3.60 g

Solution :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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