6.0 g of urea (molecular weight = 60) was dissolved in 9.9 moles of water. If the vapour pressure of pure water is `p_(0)`, the vapour pressure of solution is

Options

A. `0.10 p_(0)`
B. `1.10 p_(0)`
C. `0.90 p_(0)`
D. `0.99 p_(0)`

Answer

d

Explanation

Mole of urea `= (6)/(60) = 0.10`
According to Raoult's law
`(p_("solvent") - p_("solution"))/(p_("solvent")) = x_("solute") = (n_(A))/(n_(A) + n_(B))`
Hence, `(p_(0) - p_("solution"))/(p_(0)) = (0.10)/(0.10 + 9.9)`
or `p_(0) - p_("solution") = 0.01 p_(0)` or `p_("solution") = 0.99 p_(0)`