What will be the osmotic pressure in Pascals exerted by a solution prepared by dissolving 1 gram of polymer?

Vapour pressure of pure water at 298 K is 23.8 mm. 50 g of urea (NH2CONH2) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.

Vapoure pressure of pure water (solvent) at 298 K, p0 = 23.8 mmVapour pressure of solution, p = ?Mass of solvent ,W = 850 gMass of solute,M = 50 g

Mol. mass of water (H2O), M = 18 g mol–1
Mol.mass of urea NH2 CO NH2
= 14 + 2 + 12 + 16 + 14 + 2

                     p=p0-w×Mm×W×p°

                   p=23.8-50×1860×850

                      =23.8-0.017=23.78

Text Solution

30.96 Pa34.36 Pa68.72 Pa 48.25 Pa

Answer : B

Solution : No. of moles of polymer =`1/(150,000)``pi=CRT = n/VRT` <br> `pi=1/(150,000)xx(8.314xx10^3xx310)/0.5=34.36` Pa (R=`8.314xx10^3` Pa L `K^(-1) mol^(-1)`)

Text Solution

Answer : `30.96 Pa`

Solution : `W_(2)=1.0g, Mw_(2)=185000g mol^(-1)` <br> `T=37^(@)C=37+273=310K` <br> `R=8.314 K Pa L K ^(-1) mol^(-1)` <br> `=8.314xx10^(3) Pa L K^(-1)mol^(-1)` <br> `V=450mL=0.450 L,` osmotic pressure `(pi)=?` <br> `n_(2)=(W_(2))/(Mw_(2))=(1.0g)/(185,000g mol^(-1))=(1)/(185000)mol` <br> Applying the formula , we get <br> `pi=MRT=(n_(2))/(V(i n L))xxRT` <br> `=(1)/(185000xx0.450L)xx8.31410^(3)Pa L K^(-1)mol^(-1)xx310K` ltbr. `=30.96Pa`