एक विलेय के (2.5,g) से (250,mL) विलयन बनाया गया। (300,K) पर \(\pi=1.23,atm\) है। यदि \(i=2\), तो वास्तविक मोलर द्रव्यमान कितना होगा?
A (250,mL) solution is prepared from (2.5,g) solute. At (300,K), \(\pi=1.23,atm\). If \(i=2\), what is the true molar mass?
Correct answer and explanation
C. \(200,g,mol^{-1}\)
Concept
\(C=\frac{1.23}{2\times0.082\times300}=0.025,M\)। / \(C=\frac{1.23}{2\times0.082\times300}=0.025,M\).
Why this answer is correct
\(250,mL=0.25,L\), इसलिए मोल \(0.025\times0.25=0.00625\) हैं। / \(250,mL=0.25,L\), so moles \(=0.025\times0.25=0.00625\).
Exam Tip
मोलर द्रव्यमान \(=\frac{2.5}{0.00625}=400,g,mol^{-1}\)। / Molar mass \(=\frac{2.5}{0.00625}=400,g,mol^{-1}\).
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