यदि किसी विलेय के (1.6,g) को (200,g) जल में घोलने पर \(\Delta T_f=0.372,K\) है और \(i=2\), तो वास्तविक मोलर द्रव्यमान क्या होगा?
If (1.6,g) solute dissolved in (200,g) water gives \(\Delta T_f=0.372,K\) and \(i=2\), what is the true molar mass?
Explanation opens after your attempt
C. \(80,g,mol^{-1}\)
Concept
प्रभावी मोललता \(\frac{0.372}{1.86}=0.2\) है। / Effective molality \(=\frac{0.372}{1.86}=0.2\).
Why this answer is correct
\(i=2\), इसलिए वास्तविक मोललता (0.1) है। / Since \(i=2\), true molality is (0.1).
Exam Tip
\(200,g=0.2,kg\), मोल \(0.1\times0.2=0.02\), अतः मोलर द्रव्यमान \(=\frac{1.6}{0.02}=80,g,mol^{-1}\)। / \(200,g=0.2,kg\), moles \(=0.1\times0.2=0.02\), so molar mass \(=80,g,mol^{-1}\).
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