Question

At 27 °C, $x \, g$ of $\text{CaCl}_2$ was dissolved in 2.5 L of water. The osmotic pressure of the resultant solution is 0.82 atm. What is $x$ in grams? (Given $i = 2.5$, $R = 0.082 \, \text{L atm mol}^{-1} \text{K}^{-1}$)

• A. $37$
• B. $1.85$
• C. $3.7$
• D. $18.5$

Hint:

When given an answer, check your units and calculation methods to ensure they align with expected results, particularly in practical chemistry applications.

Answer 1

The Correct Option is C

Step 1: Use the osmotic pressure formula.

The osmotic pressure equation is: $$\Pi = i M R T$$ where: - $\Pi = 0.82$ atm (osmotic pressure), - $i = 2.5$ (Van't Hoff factor for CaCl$_2$), - $R = 0.082$ L atm mol$^{-1}$K$^{-1}$, - $T = 27^\circ C = 273 + 27 = 300$ K, - $V = 2.5$ L.

Step 2: Calculate the molarity ($M$).

Rearranging the equation: $$M = \frac{\Pi}{i R T}$$ Substituting the values: $$M = \frac{0.82}{2.5 \times 0.082 \times 300}$$ $$M = \frac{0.82}{61.5} = 0.0133 \, \text{mol/L}$$

Step 3: Calculate the moles of CaCl$_2$.

Since $M = \frac{{\text{moles of solute}}}{{\text{volume in liters}}}$, we get: $$\text{Moles of CaCl}_2 = 0.0133 \times 2.5 = 0.03325 \, \text{moles}$$

Step 4: Convert moles to grams.

Molar mass of CaCl$_2$: $$40 + 2(35.5) = 111 \, \text{g/mol}$$ Mass of CaCl$_2$ dissolved: $$x = 0.03325 \times 111 = 3.69 \approx 3.7 \, \text{g}$$

Thus, $x$ is 3.7 g.