एक विलेय के (4.5,g) से (500,mL) विलयन बना। (300,K) पर \(\pi=0.738,atm\) है। यदि विलेय (i=1.2) दिखाता है, तो वास्तविक मोलर द्रव्यमान कितना होगा?
A (500,mL) solution is prepared from (4.5,g) solute. At (300,K), \(\pi=0.738,atm\). If the solute shows (i=1.2), what is the true molar mass?
Explanation opens after your attempt
C. \(150,g,mol^{-1}\)
Concept
\(C=\frac{0.738}{1.2\times0.082\times300}=0.025,M\)। / \(C=\frac{0.738}{1.2\times0.082\times300}=0.025,M\).
Why this answer is correct
(500,mL=0.5,L), इसलिए मोल \(0.025\times0.5=0.0125\) हैं। / (500,mL=0.5,L), so moles \(=0.025\times0.5=0.0125\).
Exam Tip
मोलर द्रव्यमान \(\frac{4.5}{0.0125}=360,g,mol^{-1}\)। / Molar mass \(=\frac{4.5}{0.0125}=360,g,mol^{-1}\).
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