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natulia [17]
3 years ago
14

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
Chemistry
1 answer:
Setler79 [48]3 years ago
8 0

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

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