किसी विलेय के (2,g) को (250,g) विलायक में घोलने पर क्वथनांक उन्नयन (0.052,K) है। \(K_b=0.52,K,kg,mol^{-1}\) हो तो मोलर द्रव्यमान क्या है?
When (2,g) of a solute is dissolved in (250,g) solvent, the boiling point elevation is (0.052,K). If \(K_b=0.52,K,kg,mol^{-1}\), what is the molar mass?
Explanation opens after your attempt
C. \(80,g,mol^{-1}\)
Concept
\(m=\frac{0.052}{0.52}=0.1\) है। / \(m=\frac{0.052}{0.52}=0.1\).
Why this answer is correct
\(250,g=0.25,kg\), इसलिए मोल \(0.1\times0.25=0.025\) हैं। / \(250,g=0.25,kg\), so moles \(=0.1\times0.25=0.025\).
Exam Tip
मोलर द्रव्यमान \(=\frac{2}{0.025}=80,g,mol^{-1}\)। / Molar mass \(=\frac{2}{0.025}=80,g,mol^{-1}\).
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