यदि \(K_b=0.52,K,kg,mol^{-1}\), विलेय (1.2,g), विलायक (100,g), और मोलर द्रव्यमान \(60,g,mol^{-1}\) है, तो अवियोजित विलेय के लिए \(\Delta T_b\) कितना होगा?
If \(K_b=0.52,K,kg,mol^{-1}\), solute mass is (1.2,g), solvent mass is (100,g), and molar mass is \(60,g,mol^{-1}\), what will be \(\Delta T_b\) for a non-dissociated solute?
Correct answer and explanation
B. (0.104,K)
Concept
विलेय के मोल \(\frac{1.2}{60}=0.02\) हैं। / Moles of solute \(=\frac{1.2}{60}=0.02\).
Why this answer is correct
(100,g=0.1,kg), इसलिए मोललता (0.2,m) है। / (100,g=0.1,kg), so molality is (0.2,m).
Exam Tip
\(\Delta T_b=0.52\times0.2=0.104,K\)। / \(\Delta T_b=0.52\times0.2=0.104,K\).
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