किसी विलयन में (3,g) विलेय (250,g) विलायक में घुला है। \(K_b=0.5,K,kg,mol^{-1}\), \(\Delta T_b=0.30,K\) हो, तो मोलर द्रव्यमान कितना होगा?
A solution has (3,g) solute dissolved in (250,g) solvent. If \(K_b=0.5,K,kg,mol^{-1}\) and \(\Delta T_b=0.30,K\), what is the molar mass?
Correct answer and explanation
B. \(20,g,mol^{-1}\)
Concept
\(M_B=\frac{K_b w_B 1000}{\Delta T_b w_A}\) लगाएँ। / Use \(M_B=\frac{K_b w_B 1000}{\Delta T_b w_A}\).
Why this answer is correct
\(M_B=\frac{0.5\times3\times1000}{0.30\times250}=20,g,mol^{-1}\) होगा। / \(M_B=\frac{0.5\times3\times1000}{0.30\times250}=20,g,mol^{-1}\).
Exam Tip
विलायक के ग्राम द्रव्यमान को सूत्र में ठीक रखें। / Keep the solvent mass in grams correctly in the formula.
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