JEE Mains · Chemistry · STD 12 -1. Solution and colligative properties
A solution is prepared by dissolving 0.3 g of a non - volatile non - electrolyte solute 'A' of molar mass 60 g \( mol^{-1} \) and 0.9 g of a non - volatile non - electrolyte solute 'B' of molar mass 180 g \( mol^{-1} \) in 100 mL \( H_{2}O \) at \( 27^{\circ}C \). Osmotic pressure of the solution will be
[Given: \( R=0.082~L \) atm \( K^{-1} \) \( mol^{-1} \)]
- A 1.23 atm
- B 2.46 atm
- C 0.82 atm
- D 1.47 atm
Answer & Solution
Correct Answer
(B) 2.46 atm
Step-by-step Solution
Detailed explanation
Mass of solute 'A' = 0.3 g Moles of solute 'A' = \( \frac{0.3g}{60g/mol} = \frac{1}{200} \) mol Mass of solute 'B' = 0.9 g Moles of solute 'B' = \( \frac{0.9g}{180g/mol} = \frac{1}{200} \) mol Total molarity of all solutes = \( \frac{2/200}{100} \times 1000 = \frac{1}{10}M \)…
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