Question

Calculate osmotic pressure exerted by a solution containing $0.822\ \mathrm{g}$ of solute in $300\ \mathrm{mL}$ of water at $300\ \mathrm{K}$. (Molar mass of solute = $340\ \mathrm{g\ mol^{-1}}$, $R = 0.0821\ \mathrm{L\ atm\ mol^{-1}\ K^{-1}}$)

• A. $0.5\ \mathrm{atm}$
• B. $0.2\ \mathrm{atm}$
• C. $0.1\ \mathrm{atm}$
• D. $0.4\ \mathrm{atm}$

Hint:

To simplify calculations under pressure, group terms together: $\frac{300}{0.3} = 1000$. Then $1000 \times 0.822 = 822$. This quickly reduces messy fractions into manageable mental math!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the osmotic pressure ($\pi$) of a solution containing a known mass of solute ($W_2 = 0.822\ \mathrm{g}$) dissolved in a specific volume of water ($V = 300\ \mathrm{mL}$) at a given temperature ($T = 300\ \mathrm{K}$).

Step 2: Key Formula or Approach:
The formula for osmotic pressure is: $$\pi = CRT = \frac{W_2 \cdot R \cdot T}{M_2 \cdot V}$$ Where $M_2$ is the molar mass of the solute and $V$ is the volume of the solution in liters.

Step 3: Detailed Explanation:
Let's list the given parameters and convert units where necessary:

• Mass of solute ($W_2$) = $0.822\ \mathrm{g}$

• Molar mass ($M_2$) = $340\ \mathrm{g\ mol^{-1}}$

• Volume ($V$) = $300\ \mathrm{mL} = 0.3\ \mathrm{L}$

• Temperature ($T$) = $300\ \mathrm{K}$

• Gas Constant ($R$) = $0.0821\ \mathrm{L\ atm\ mol^{-1}\ K^{-1}}$
Substitute these values into the osmotic pressure equation: $$\pi = \frac{0.822 \times 0.0821 \times 300}{340 \times 0.3}$$ Simplify the expression by canceling terms: $$\pi = \frac{0.822 \times 0.0821 \times 1000}{340}$$ $$\pi = \frac{822 \times 0.0821}{340} = \frac{67.4862}{340} \approx 0.1985\ \mathrm{atm} \approx 0.2\ \mathrm{atm}$$

Step 4: Final Answer:
The osmotic pressure exerted by the solution is $0.2\ \mathrm{atm}$, corresponding to option (B).