एक विलेय के (1.8,g) से (300,mL) विलयन बना। (300,K) पर \(\pi=0.492,atm\) है। यदि (i=1.2), तो वास्तविक मोलर द्रव्यमान कितना होगा?
A (300,mL) solution is prepared from (1.8,g) solute. At (300,K), \(\pi=0.492,atm\). If (i=1.2), what is the true molar mass?
Correct answer and explanation
C. \(150,g,mol^{-1}\)
Concept
\(C=\frac{0.492}{1.2\times0.082\times300}=0.0167,M\)। / \(C=\frac{0.492}{1.2\times0.082\times300}=0.0167,M\).
Why this answer is correct
(300,mL=0.3,L), इसलिए मोल \(0.0167\times0.3=0.005\) हैं। / (300,mL=0.3,L), so moles \(=0.0167\times0.3=0.005\).
Exam Tip
मोलर द्रव्यमान \(=\frac{1.8}{0.005}=360,g,mol^{-1}\)। / Molar mass \(=\frac{1.8}{0.005}=360,g,mol^{-1}\).
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