When 1-80 gm glucose dissolved in 90 gm of H 2 O- the mole fraction of glucose isA- 0-00399B- 0-00199C- 0-0199D- 0-998

The correct option is B 0.00199x_Glucose=M/n_glu+n_water=0.01/0.01+5=0.00199 ...

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Percentage Composition

When 1.80 gm ...

Question

When 1.80 gm glucose dissolved in 90 gm of $H_{2} O$ , the mole fraction of glucose is

0.00399

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0.00199

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0.0199

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0.998

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