परासरण दाब विधि में (1.5,g) विलेय (500,mL) विलयन में है। \(T=300,K\), \(\pi=0.738,atm\), \(R=0.082,L,atm,K^{-1},mol^{-1}\) हो, तो मोलर द्रव्यमान कितना होगा?
In osmotic pressure method, (1.5,g) solute is present in (500,mL) solution. If \(T=300,K\), \(\pi=0.738,atm\), and \(R=0.082,L,atm,K^{-1},mol^{-1}\), what is the molar mass?
Explanation opens after your attempt
C. \(100,g,mol^{-1}\)
Concept
आयतन \(500,mL=0.5,L\) करें और \(M=\frac{wRT}{\pi V}\) लगाएँ। / Convert \(500,mL=0.5,L\) and use \(M=\frac{wRT}{\pi V}\).
Why this answer is correct
\(M=\frac{1.5\times0.082\times300}{0.738\times0.5}=100,g,mol^{-1}\) होगा। / \(M=\frac{1.5\times0.082\times300}{0.738\times0.5}=100,g,mol^{-1}\).
Exam Tip
मिलीलीटर को लीटर में बदलना न भूलें। / Do not forget to convert millilitres into litres.
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