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

Answer

`1.2xx10^(21)` molecules

Explanation

`because DeltaT=(100xxK_(b)xxw)/(mxxW)`
`therefore 0.01=(1000xx0.5xxw)/(180xx100)`
`therefore w=0.36g`
`therefore` No. of molecules of glucose in `0.36g`
`=(6.023xx10^(23)xx0.36)//180`
`=1.2xx10^(21)` molecules