यदि (1.0,g) विलेय (100,g) विलायक में है और \(K_f=1.86,K,kg,mol^{-1}\), मोलर द्रव्यमान \(100,g,mol^{-1}\) है, तो आदर्श \(\Delta T_f\) कितना होगा?
If (1.0,g) solute is present in (100,g) solvent and \(K_f=1.86,K,kg,mol^{-1}\), with molar mass \(100,g,mol^{-1}\), what is the ideal \(\Delta T_f\)?
Explanation opens after your attempt
B. (0.186,K)
Concept
विलेय के मोल \(\frac{1.0}{100}=0.01\) हैं और विलायक (0.100,kg) है। / Moles of solute are \(\frac{1.0}{100}=0.01\) and solvent is (0.100,kg).
Why this answer is correct
मोललता \(m=\frac{0.01}{0.100}=0.1\) है, इसलिए \(\Delta T_f=1.86\times0.1=0.186,K\) है। / Molality \(m=\frac{0.01}{0.100}=0.1\), so \(\Delta T_f=1.86\times0.1=0.186,K\).
Exam Tip
पहले मोललता निकालें, फिर अवनमन निकालें। / First find molality, then depression.
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