यदि \(\Delta T_b=0.312,K\), \(K_b=0.52,K,kg,mol^{-1}\), विलायक (100,g) और विलेय (3,g) है, तो अवियोजित विलेय का मोलर द्रव्यमान क्या है?
If \(\Delta T_b=0.312,K\), \(K_b=0.52,K,kg,mol^{-1}\), solvent mass is (100,g), and solute mass is (3,g), what is the molar mass of the non-dissociated solute?
Correct answer and explanation
B. \(50,g,mol^{-1}\)
Concept
\(m=\frac{0.312}{0.52}=0.6\)। / \(m=\frac{0.312}{0.52}=0.6\).
Why this answer is correct
\(100,g=0.1,kg\), इसलिए मोल \(0.6\times0.1=0.06\) हैं। / \(100,g=0.1,kg\), so moles \(=0.6\times0.1=0.06\).
Exam Tip
मोलर द्रव्यमान \(=\frac{3}{0.06}=50,g,mol^{-1}\)। / Molar mass \(=\frac{3}{0.06}=50,g,mol^{-1}\).
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