किसी अवियोजित विलेय के (4,g) को (200,g) विलायक में घोलने पर \(\Delta T_b=0.104,K\) है। \(K_b=0.52,K,kg,mol^{-1}\) हो तो मोलर द्रव्यमान क्या होगा?
When (4,g) of a non-dissociated solute is dissolved in (200,g) solvent, \(\Delta T_b=0.104,K\). If \(K_b=0.52,K,kg,mol^{-1}\), what is the molar mass?
Explanation opens after your attempt
B. \(100,g,mol^{-1}\)
Concept
\(\Delta T_b=K_bm\), so \(m=\frac{0.104}{0.52}=0.2\).
Why this answer is correct
(200,g=0.2,kg), so moles \(=0.2\times0.2=0.04\).
Exam Tip
Molar mass \(=\frac{4}{0.04}=100,g,mol^{-1}\). चरण 1: \(\Delta T_b=K_bm\), इसलिए \(m=\frac{0.104}{0.52}=0.2\)। चरण 2: (200,g=0.2,kg), इसलिए मोल \(0.2\times0.2=0.04\) हैं। चरण 3: मोलर द्रव्यमान \(=\frac{4}{0.04}=100,g,mol^{-1}\)।
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