9 gram anhydrous oxalic acid (mol. wt. = 90) was dissolved in 9.9 moles of water. If vapour pressure of pure water is `P_(1)^(@)` the vapour pressure of solution is

Options

A. `0.99 p_(1)^(@)`
B. `0.1 p_(1)^(@)`
C. `0.99 p_(1)^(@)`
D. `1.1 p_(1)^(@)`

Answer

A

Explanation

The total vapour pressure of a solution in this case only depends on vapour pressure of water as anhydrous oxalic acid is a non-volatile compound.
`:.` Vapour pressure of solution = vapour pressure of water `(p_(w))`
According to Raoult law
`P_(w) = x_(w) P_(w)^(@)`
Number of mole of oxalic acid `= (9)/(90) = 0.1` mole
`:. X_(w) = (9.9)/(9.9 + 0.1) = 0.99`
`implies P_(s) = p_(w) = 0.99 xx P_(1)^(@)`