Question

6.8g of a compound is dissolved in 100g of water. Calculate osmotic pressure of this solution at 298K when the boiling point of the solution is 100.11℃. Given Kb= 0.52K/m and R= 0.0821 L atm/K

Answer 1

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Answer 2

Answer:

The correct answer is 5.169 atm.

Explanation:

Osmotic pressure of a solution is the least or minimum pressure that is to be applied to a solution to prevent its osmosis.

According to the question:

Weight of compound   $\left(\mathrm{W}_{\mathrm{A}}\right)$ = 6.8 g      

Weight of water  $\left(\mathrm{W}_{\mathrm{B}}\right)$  = 100 g                

Temperature (T) = 298 K

Boiling Point of solution  = 100.11 0C = 373.11 K      

$\Delta T_{b}$ = 0.11 K                                            

$\mathrm{K}_{\mathrm{b}}$ = 0.52K/m                                

R= 0.0821 L atm/K

Thus, $M_{B}$ =  $\left(\mathrm{W}_{\mathrm{A}} \times \mathrm{K}_{\mathrm{b}} \times 1000\right)/ \left(\Delta T_{\mathrm{b}} \times \mathrm{W}_{\mathrm{A}}\right)$  

$=\frac{6.8 \times 0.52 \times 1000}{0.11 \times 100}= 321.4545 g/mol$                              

Thus Osmotic pressure (π) =                                      

$=\frac{6.8 \times 0.52 \times 1000}{0.11 \times 100}=321.4545 \mathrm{g} / \mathrm{mol}$

       = 5.169 atm.