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 is

Options

A. `6.023xx10^(23)`
B. `6.023xx10^(22)`
C. `12.046xx10^(20)`
D. `12.046xx10^(23)`

Answer

C

Explanation

`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)`