Answer: e. 3.5 atm
Explanation:
[tex]\pi =CRT[/tex]
[tex]\pi[/tex] = osmotic pressure = ?
C= concentration in Molarity
R= solution constant = 0.0821 Latm/Kmol
T= temperature = [tex]36^0C=(36+273)K=309K[/tex]
For the given solution: 0.82 grams of [tex]NaCl[/tex] is dissolved in 100 g of solution.
[tex]{\text {volume of solution}}=\frac{\text {mass of solution}}{\text {Density of solution}}=\frac{100g}{1.0g/ml}=100ml[/tex]
[tex]Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution in ml}}[/tex]
Putting in the values we get:
[tex]C_{NaCl}=\frac{0.82\times 1000}{58.5\times 100}=0.14M[/tex]
[tex]\pi=0.14mol/L\times 0.0821Latm/Kmol\times 309K[/tex]
[tex]\pi=3.5atm[/tex]
The osmotic pressure of a 0.82% NaCl by weight aqueous solution is 3.5 atm
