किसी विलेय के (3,g) को (250,g) जल में घोलने पर \(\Delta T_f=0.465,K\) है। यदि (i=1.5), तो वास्तविक मोलर द्रव्यमान कितना होगा?
When (3,g) solute is dissolved in (250,g) water, \(\Delta T_f=0.465,K\). If (i=1.5), what is the true molar mass?
Explanation opens after your attempt
B. \(72,g,mol^{-1}\)
Concept
प्रभावी मोललता \(\frac{0.465}{1.86}=0.25\) है। / Effective molality \(=\frac{0.465}{1.86}=0.25\).
Why this answer is correct
वास्तविक मोललता \(\frac{0.25}{1.5}=0.1667\) होगी। / True molality \(=\frac{0.25}{1.5}=0.1667\).
Exam Tip
(250,g=0.25,kg), मोल \(0.1667\times0.25=0.0417\), अतः मोलर द्रव्यमान \(\frac{3}{0.0417}\approx72,g,mol^{-1}\)। / (250,g=0.25,kg), moles \(=0.1667\times0.25=0.0417\), so molar mass \(=\frac{3}{0.0417}\approx72,g,mol^{-1}\).
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