Question

At \(300\,K\), \(x\) moles of \(CaCl_2\) \((i=2.5;\ \text{molar mass}=111\,g\,mol^{-1})\) is dissolved in \(2.5\,L\) of water. The osmotic pressure of the resultant solution is \(0.75\,atm\). What is \(\Delta T_b\) of the solution? \[ (\text{density of water}=1\,g\,mL^{-1},\; K_b=0.52\,K\,kg\,mol^{-1},\; R=0.08\,L\,atm\,mol^{-1}K^{-1}) \]

• A. \(0.016\,K\)
• B. \(0.032\,K\)
• C. \(0.048\,K\)
• D. \(0.064\,K\)

Hint:

Use osmotic pressure first to find concentration: \[ \pi=iCRT \] Then use \[ \Delta T_b=iK_bm. \] For dilute aqueous solutions, \[ 1\,L\ \text{water}\approx1\,kg\ \text{water} \] when density is \(1\,g\,mL^{-1}\).

Answer 1

The Correct Option is A

Concept: Osmotic pressure is given by \[ \pi=iCRT \] where \[ C=\frac{n}{V}. \] Elevation in boiling point: \[ \Delta T_b=iK_bm. \]

Step 1: Calculate the molarity of the solution. Given, \[ \pi=0.75\ atm, \qquad i=2.5, \qquad T=300\ K, \qquad R=0.08. \] Using \[ \pi=iCRT, \] \[ 0.75=(2.5)C(0.08)(300). \] \[ 0.75=60C. \] \[ C=\frac{0.75}{60} =0.0125\ M. \]

Step 2: Calculate moles of \(CaCl_2\). Volume of solution \(\approx 2.5\,L\) \[ n=CV \] \[ =(0.0125)(2.5) \] \[ =0.03125\ mol. \]

Step 3: Calculate molality. Mass of water: \[ 2.5\,L = 2500\,mL. \] Since density of water is \[ 1\,g\,mL^{-1}, \] \[ \text{mass of water}=2500\,g=2.5\,kg. \] Therefore, \[ m=\frac{0.03125}{2.5} \] \[ =0.0125\ mol\,kg^{-1}. \]

Step 4: Calculate elevation in boiling point. \[ \Delta T_b=iK_bm. \] \[ =(2.5)(0.52)(0.0125). \] \[ =0.01625\ K. \] \[ \Delta T_b\approx0.016\ K. \]

Final Answer: \[ \boxed{\Delta T_b=0.016\ K} \] \[ \boxed{\text{Answer = (A)}} \]