एक विलेय के (1.2,g) से (200,mL) विलयन बनाया गया। (300,K) पर \(\pi=0.246,atm\) है। यदि विलेय द्विमर बना रहा हो और (i=0.5) हो, तो वास्तविक मोलर द्रव्यमान क्या होगा?
A (200,mL) solution is prepared from (1.2,g) solute. At (300,K), \(\pi=0.246,atm\). If the solute forms dimers and (i=0.5), what is the true molar mass?
Explanation opens after your attempt
C. \(240,g,mol^{-1}\)
Concept
\(\pi=iCRT\), so \(C=\frac{0.246}{0.5\times0.082\times300}=0.02,M\).
Why this answer is correct
(200,mL=0.2,L), so moles \(=0.02\times0.2=0.004\).
Exam Tip
True molar mass \(=\frac{1.2}{0.004}=300,g,mol^{-1}\). चरण 1: \(\pi=iCRT\), इसलिए \(C=\frac{0.246}{0.5\times0.082\times300}=0.02,M\)। चरण 2: (200,mL=0.2,L), अतः मोल \(0.02\times0.2=0.004\) हैं। चरण 3: वास्तविक मोलर द्रव्यमान \(=\frac{1.2}{0.004}=300,g,mol^{-1}\)।
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