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An aqueous solution of glucose boils at `100.01^(@)C`.The molal elevation constant for water is `0.5 kmol^(-1)kg`. The number of molecules of glucose in the solution containing `100g` of water isA. `6.023xx10^(23)`B. `6.023xx10^(22)`C. `12.046xx10^(20)`D. `12.046xx10^(23)` |
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Answer» Correct Answer - C `Delta"T"_(b)="K"_(b).m` Or `"m"=(DeltaT_(b))/(K_(b))=0.01/0.5=.02 " mole Kg" ^(-1) "of water"` lt brgt So, the number of moles of glucose in 100g of water `=(0.02xx100)/1000=0.002` mole of glucose `=0.002xx6.023xx10^(23)=2xx6.023xx10^(20)` |
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