Question

A solution obtained by mixing $100mL$ of $20\%$ solution of urea ($molarmass=60$) and $10mL$ of $1.65$ solution of cane sugar ($molarmass=342$) at $300K$. $R=0.083Lbar{K}^{-1}mo{l}^{-1}$). Calculate.
(a)Osmotic pressure of urea solution, (b) Osmotic pressure of cane sugar solution, (c) Total osmotic pressure of solution:

Answer 1

$\Pi =CRT$

Number of moles of urea$=\frac{20}{60}=\frac{1}{3}=0.33$ mole

Number of moles of cane$=\frac{1.65}{342}=0.0047$ mole

(i) Concentration of urea in $100ml=\frac{0.33\times 1000}{100}=3.3M$

Osmotic pressure of urea$=3.3\times 0.08257\times 300=81.7443atm$

(ii) Osmotic pressure of cane sugar$=\frac{0.0047\times 1000}{10}=0.47M$

Osmotic pressure$=0.47\times 0.08257\times 300=11.642atm$

(iii) Osmotic pressure of solution$=0.0047+0.33=0.3347$ moles

Concentration of solute in $110ml$ solution$=\frac{0.3347\times 1000}{110}=33.47M$

Osmotic pressure of solution= $33.47\times 0.08257\times 300=829.08atm$ .