If the vapor pressure of a dilute aqueous solution of glucose is `750 mm` of Hg at `373 K`, then molality of solute is

Options

A. `0.64`
B. `0.741`
C. `0.68`
D. `0.94`

Answer

B

Explanation

`(P^(@)-P_(s))/(P_(s))=(W_(1)xxm_(2))/(m_(1)xxw_(2))`
`rArr (P^(@)-P_(s))/(P_(s))=((w_(1)xx1000))/((m_(1)xxw_(2)))xx(m_(2))/(1000)=mxx(m_(2))/(1000) " " (m_(1)`= mol. Wt. of solute)
`rArr (760-750)/(750)=mxx(18)/(1000)`
`:.m=(10xx1000)/(750xx18) =0.741` mol/ k of solvent
Hence, (B) is the correct answer.