Question

what is the osmotic pressure of a 0.25 l solution that contains 0.88 g of naphthalene (c₁₀h₈) at 35.00°c?

• use -273.15°c for absolute zero.
• use r = 0.08206 l atm/mol k for the ideal gas constant.

select the correct answer below:
0.37 atm
0.58 atm
1.7 atm
0.69 atm

Explanation:

Step1: Calculate the molar mass of naphthalene ($C_{10}H_8$)

The molar mass of $C$ is $12.01\ g/mol$ and of $H$ is $1.008\ g/mol$.
For $C_{10}H_8$: $M=(10\times12.01)+(8\times1.008)=120.1 + 8.064=128.164\ g/mol$

Step2: Calculate the number of moles ($n$) of naphthalene

Using the formula $n=\frac{m}{M}$, where $m = 0.88\ g$ and $M = 128.164\ g/mol$
$n=\frac{0.88\ g}{128.164\ g/mol}\approx0.00687\ mol$

Step3: Convert temperature to Kelvin

Using the formula $T=t + 273.15$, where $t = 35.00^{\circ}C$
$T=35.00+273.15 = 308.15\ K$

Step4: Calculate the osmotic pressure ($\pi$)

The formula for osmotic pressure is $\pi=\frac{nRT}{V}$
Substitute $n = 0.00687\ mol$, $R=0.08206\ \frac{L\ atm}{mol\ K}$, $T = 308.15\ K$ and $V=0.25\ L$
$\pi=\frac{0.00687\times0.08206\times308.15}{0.25}$
First calculate the numerator: $0.00687\times0.08206\times308.15\approx0.00687\times25.28\approx0.173$
Then $\pi=\frac{0.173}{0.25}=0.692\ atm\approx0.69\ atm$

Answer:

0.69 atm