Question

Calculate the osmotic pressure of a solution obtained by mixing 100cm^3 of 1.5% solution of urea (mol. Mass =60) and 100cm^3 of 3.42% solution of sugarcane (mol.mass= 342) at 20degree celcius (R= 0.082 litreatm/degree/mole) ?

Answer 1

Answer:

and 100cm3 of 3 42 % solution of cane sugar (molar mass=342) at 20oC (R=0 ... by mixing 100cm3 of 1.5% solution of urea (molar mass =60) and 100cm3 of 3.42 ... Moles of cane sugar in 100 cm3 of solution = 3.42/342.

Explanation:

Answer 2

Answer:

$\boxed{4.20\:atm}$

Explanation:

Moles of urea in 100 cm3 of solution = $\frac{1.5}{60}$

Moles of cane sugar in 100 cm3 of solution = $\frac{3.42}{342}$

Total Moles = $\frac{1.5}{60} +\frac{3.42}{342}$

= 0.035 in 200 cm3 of solution

As osmotic pressure is a colligative property it will depend upon total moles of solute present in solution and not upon their nature

Hence,

$\pi V=nRT$

$\pi=\frac{nRT}{V}$

$=\frac{0.035X0.082}{200/1000}$

$=\frac{0.035X0.082X293X1000}{200}$

$\pi =4.20atm$