The vapore pressure of 100 g of water reduces from 17.53 mm to 17.22 mm when 17.10 g of substance 'X' is dissolved in it. Substance X can be

Options

A. methanol
B. glucose
C. carbon dioxide
D. cannot predict

Answer

B

Explanation

`"Given, " p^(@)=17.53,p_(s)=17.22andw=17.10`
`(p^(@)-p_(s))/(p^(@))=(n)/(n+N)=((w)/(m))/((w)/(m)+(W)/(M))`
`:.(w)/(m)lt lt lt(W)/(M)`
`:.(p^(@)-p_(s))/(p^(@))=(w//m)/(W//M)=(w)/(m)xx(M)/(W)`
`rArr(17.53-17.22)/(17.53)=(17.10)/(m)xx(18)/(100)`
`rArrm=(17.10xx18xx17.53)/(0.31xx100)=174.05`
174 is nearest to molecular weight of glucose `(C_(6)H_(12)O_(6)),` thus the substance 'X' can be glucose.