Question

The osmotic pressure of a solution formed by dissolving 35.0 mg of aspirin (c9h8o4) in 0.250 l of water at 25°c is __________ atm.

Answer 1

Answer: The osmotic pressure of aspirin is 0.0190 atm

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

We are given:

Mass of solute (aspirin) = 35.0 mg = 0.035 g    (conversion factor: 1 g = 1000 mg)

Molar mass of aspirin = 180.16 g/mol

Volume of solution = 0.250 L

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{0.035g}{180.16g/mol\times 0.250L}\\\\\text{Molarity of solution}=7.77\times 10^{-4}M$

To calculate the osmotic pressure, we use the equation:

$\pi=iMRT$

where,

$\pi$ = osmotic pressure of the solution

i = Van't hoff factor = 1 (for non-electrolytes)

M = molarity of aspirin = $7.77\times 10^{-4}M$

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$\pi=1\times 7.77\times 10^{-4}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 298K\\\\\pi=0.0190atm$

Hence, the osmotic pressure of aspirin is 0.0190 atm

Answer 2

Mass of aspirin = 0.025 g
Molar mass of C9H8O4 is 180.1583 g/mol
moles of aspirin = .025g / 180.1583 g/mol = 0.000138767 moles
volume solution = .250 L
molarity of the solution = 0.000138767 moles / .250L =5.551 x 10 ^-04 Moles / liter
for aspirin i = Vant'Hoff factor = 1 particle in solution
T = 25 + 273 =298 K
osmotic pressure = M x R x T x i =
5.551 x 10 ^-04 mole L -1 x 0.08206 L atm K−1 mol−1 x 298 K x 1 = 0.0136 atmospheres