Question

9 gram anhydrous oxalic acid (mol. wt. = 90) was dissolved in 9.9 moles of water. If vapour pressure of pure water is pf the vapour pressure of solution is ______.

पर्याय

• 0.99 \[\ce{p^\circ_1}\]

• 0.1 \[\ce{p^\circ_1}\]

• 0.99 \[\ce{p^\circ_2}\]

• 1.1 \[\ce{p^\circ_1}\]

Answer 1

9 gram anhydrous oxalic acid (mol. wt. = 90) was dissolved in 9.9 moles of water. If vapour pressure of pure water is pf the vapour pressure of solution is 0.99 \[\ce{p^\circ_1}\].

Explanation:

Anhydrous oxalic acid is a non-volatile chemical, the total vapour pressure of a solution in this circumstance is solely determined by the vapour pressure of water.

∴ Vapour pressure of solution = vapour pressure of water (p_w)

According to Raoult's law

p_w = x_w p°_w 

Number of moles of oxalic acid = `9/90` = 0.1 moles

∴ `x_"w" = 9.9/(9.9 + 0.1) = 0.99`

`=> "p"_"s" = "p"_"w" = 0.99 xx "p"_1^circ`