Osmotic pressure of solution containing 0.6 g urea and 3.42 g sugar in 100 ml at $\text{27°C}$

Step 1: Molar masses

• Urea (NH₂CONH₂): 60 g·mol⁻¹
• Sugar (Sucrose, C₁₂H₂₂O₁₁): 342 g·mol⁻¹

Step 2: Moles of solute

• Urea: n = 0.6 / 60 = 0.01 mol
• Sugar: n = 3.42 / 342 = 0.01 mol
• Total moles: 0.01 + 0.01 = 0.02 mol

Step 3: Molarity of solution

Volume = 100 mL = 0.1 L

C = n / V = 0.02 / 0.1 = 0.2 M

Step 4: Osmotic pressure formula

π = C R T

• C = 0.2 mol·L⁻¹
• R = 0.0821 L·atm·mol⁻¹·K⁻¹
• T = 27 + 273 = 300 K

Step 5: Substitution

π = 0.2 × 0.0821 × 300

π = 4.9263 ≈ 4.92 atm

Final Answer:b) 4.92 atm