किसी विलेय के (2.4,g) को (200,g) विलायक में घोलने पर \(\Delta T_b=0.156,K\) है। यदि \(K_b=0.52,K,kg,mol^{-1}\) और (i=1.5), तो वास्तविक मोलर द्रव्यमान क्या होगा?
When (2.4,g) solute is dissolved in (200,g) solvent, \(\Delta T_b=0.156,K\). If \(K_b=0.52,K,kg,mol^{-1}\) and (i=1.5), what is the true molar mass?
Explanation opens after your attempt
C. \(60,g,mol^{-1}\)
Concept
\(m=\frac{0.156}{1.5\times0.52}=0.2\)। / \(m=\frac{0.156}{1.5\times0.52}=0.2\).
Why this answer is correct
(200,g=0.2,kg), इसलिए मोल \(0.2\times0.2=0.04\) हैं। / (200,g=0.2,kg), so moles \(=0.2\times0.2=0.04\).
Exam Tip
मोलर द्रव्यमान \(=\frac{2.4}{0.04}=60,g,mol^{-1}\)। / Molar mass \(=\frac{2.4}{0.04}=60,g,mol^{-1}\).
Login to save your score, XP, coins and progress.
