Question

At 36°C, what is the osmotic pressure of a 0.82% NaCl by weight aqueous solution? Assume the density of the solution is 1.0 g/mL. (R = 0.0821 L · atm/(K · mol)) a. 7.1 atm b. 0.35 atm c. 0.82 atm d. 4.1 × 102 atm e. 3.5 atm

Answer 1

Answer: e. 3.5 atm

Explanation:

$\pi =CRT$

$\pi$ = osmotic pressure  = ?

C= concentration in Molarity

R= solution constant  = 0.0821 Latm/Kmol

T= temperature = $36^0C=(36+273)K=309K$

For the given solution: 0.82 grams of $NaCl$ is dissolved in 100 g of solution.

${\text {volume of solution}}=\frac{\text {mass of solution}}{\text {Density of solution}}=\frac{100g}{1.0g/ml}=100ml$

$Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution in ml}}$

Putting in the values we get:

$C_{NaCl}=\frac{0.82\times 1000}{58.5\times 100}=0.14M$

$\pi=0.14mol/L\times 0.0821Latm/Kmol\times 309K$

$\pi=3.5atm$

The osmotic pressure of a 0.82% NaCl by weight aqueous solution is 3.5 atm