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

The osmotic pressure of a 0.82% NaCl solution at 36°C is 34 atm.

Step-by-step explanation:

The osmotic pressure of a solution can be calculated using the formula II = MRT, where II is the osmotic pressure, M is the molarity of the solution, R is the ideal gas constant, and T is the temperature in Kelvin.

In this case, the molarity of the solution is 0.70 M NaCl. Since 1 mol of NaCl produces 2 mol of particles, the total concentration of dissolved particles is (2)(0.70 M) = 1.4 M.

Plugging in the values into the formula, II = (1.4 mol/L) [0.0821 (L· atm)/(K · mol)] (298 K) = 34 atm.

Answer 2

Answer: e. 3.5 atm

Step-by-step 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}}=(100g)/(1.0g/ml)=100ml`

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

Putting in the values we get:

`C_(NaCl)=(0.82* 1000)/(58.5* 100)=0.14M`

`\pi=0.14mol/L* 0.0821Latm/Kmol* 309K`

`\pi=3.5atm`

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