For a silute solution conatining `2.5 g` of a non-volatile non-electrolyte solute in 100g of water, the elevation in boiling point at 1 atm pressure is `2^(@)C`. Assuming concentration of solute is much lower than the concentration (take `K_(b) = 0.76 K kg mol^(-1))`

Options

A. 724
B. 740
C. 736
D. 718

Answer

A

Explanation

From Raoult law
`(p^@ - p)/p^@ = ("No. of moles of solute")/("No. of moles of solvent + No. of moles of solute")`
When the concentration of solute is much lower than the concentration of solvent,
`(p^@ - p)/(p^@) = ("No. of moles of solute")/("No. of moles of solvent")`
`DeltaT_b = K_b xxm`
`m = ("Number of moles of the solute")/("Mass of solvent in grams")xx1000`
`DeltaT_b = K_b xx ("Number of moles of the solute")/("Mass of solvent in grams")xx1000`
Number of moles of solute
`= (DeltaT_bxx" Mass of solvent in grams")/(K_b xx 1000)`
`=(2xx100)/(0.76 xx 1000) = 0.26,`
Number of moles of solvent `= 100/18 = 5.56`
From equation (i) we get, `(760-p)/(760) = 0.26/5.56`
On solving, `p = 724.46~~724`