किसी विलेय के (4.8,g) को (400,g) विलायक में घोलने पर \(\Delta T_b=0.156,K\) है। यदि \(K_b=0.52,K,kg,mol^{-1}\) और (i=0.75), तो वास्तविक मोलर द्रव्यमान क्या होगा?
When (4.8,g) solute is dissolved in (400,g) solvent, \(\Delta T_b=0.156,K\). If \(K_b=0.52,K,kg,mol^{-1}\) and (i=0.75), what is the true molar mass?
Explanation opens after your attempt
C. \(60,g,mol^{-1}\)
Concept
\(\Delta T_b=iK_bm\), इसलिए \(m=\frac{0.156}{0.75\times0.52}=0.4\)। / From \(\Delta T_b=iK_bm\), \(m=\frac{0.156}{0.75\times0.52}=0.4\).
Why this answer is correct
(400,g=0.4,kg), इसलिए मोल \(0.4\times0.4=0.16\) हैं। / (400,g=0.4,kg), so moles \(=0.4\times0.4=0.16\).
Exam Tip
मोलर द्रव्यमान \(=\frac{4.8}{0.16}=30,g,mol^{-1}\)। / Molar mass \(=\frac{4.8}{0.16}=30,g,mol^{-1}\).
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