यदि \(K_f=1.86,K,kg,mol^{-1}\), विलेय (4,g), विलायक (200,g), मोलर द्रव्यमान \(100,g,mol^{-1}\), और (i=1.5) है, तो \(\Delta T_f\) कितना होगा?
If \(K_f=1.86,K,kg,mol^{-1}\), solute mass is (4,g), solvent mass is (200,g), molar mass is \(100,g,mol^{-1}\), and (i=1.5), what is \(\Delta T_f\)?
Explanation opens after your attempt
D. (0.558,K)
Concept
विलेय के मोल \(\frac{4}{100}=0.04\) हैं। / Moles of solute \(=\frac{4}{100}=0.04\).
Why this answer is correct
(200,g=0.2,kg), इसलिए मोललता (0.2,m) है। / (200,g=0.2,kg), so molality is (0.2,m).
Exam Tip
\(\Delta T_f=iK_fm=1.5\times1.86\times0.2=0.558,K\)। / \(\Delta T_f=iK_fm=1.5\times1.86\times0.2=0.558,K\).
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