किसी विलेय के (2.0,g) से (1,L) विलयन बना। (300,K) पर \(\pi=0.164,atm\) है। यदि विलेय (50%) द्विमर बनाता है, तो वास्तविक मोलर द्रव्यमान क्या होगा?
A (1,L) solution is prepared from (2.0,g) solute. At (300,K), \(\pi=0.164,atm\). If the solute forms dimers to the extent of (50%), what is the true molar mass?
Explanation opens after your attempt
C. \(300,g,mol^{-1}\)
Concept
(50%) द्विमर के लिए \(i=1-\frac{0.5}{2}=0.75\)। / For (50%) dimerization, \(i=1-\frac{0.5}{2}=0.75\).
Why this answer is correct
\(C=\frac{0.164}{0.75\times0.082\times300}\approx0.00889,M\)। / \(C=\frac{0.164}{0.75\times0.082\times300}\approx0.00889,M\).
Exam Tip
(1,L) में मोल (0.00889), इसलिए मोलर द्रव्यमान \(\frac{2.0}{0.00889}\approx225,g,mol^{-1}\)। / In (1,L), moles are (0.00889), so molar mass \(=\frac{2.0}{0.00889}\approx225,g,mol^{-1}\).
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