1.

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

Answer»

`0.10 p_(0)`
`1.10 p_(0)`
`0.90 p_(0)`
`0.99 p_(0)`

Solution :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)`


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