किसी विलेय के (2,g) से (500,mL) विलयन बनाया गया। (300,K) पर \(\pi=0.615,atm\) है। यदि (i=1.25), तो वास्तविक मोलर द्रव्यमान कितना होगा?
A (500,mL) solution is prepared from (2,g) solute. At (300,K), \(\pi=0.615,atm\). If (i=1.25), what is the true molar mass?
Explanation opens after your attempt
B. \(160,g,mol^{-1}\)
Concept
\(C=\frac{0.615}{1.25\times0.082\times300}=0.02,M\)। / \(C=\frac{0.615}{1.25\times0.082\times300}=0.02,M\).
Why this answer is correct
(500,mL=0.5,L), इसलिए मोल \(0.02\times0.5=0.01\) हैं। / (500,mL=0.5,L), so moles \(=0.02\times0.5=0.01\).
Exam Tip
मोलर द्रव्यमान \(=\frac{2}{0.01}=200,g,mol^{-1}\)। / Molar mass \(=\frac{2}{0.01}=200,g,mol^{-1}\).
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