The boiling point (in {}^{circ}C) of 0.1 , mo | vedclass

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(C) The boiling point elevation is given by the formula $\Delta T_{b} = i 	imes K_{b} 	imes m$. For $CuSO_{4} \cdot 5H_{2}O$,the van't Hoff factor $

Vedclass · Accepted Answer

$100.10$

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(C) The boiling point elevation is given by the formula $\Delta T_{b} = i 	imes K_{b} 	imes m$. For $CuSO_{4} \cdot 5H_{2}O$,the van't Hoff factor $i = 2$ because it dissociates as $CuSO_{4} ightarrow Cu^{2+} + SO_{4}^{2-}$. Given $K_{b} = 0.512 \, K \, kg \, mol^{-1}$ and molality $m = 0.1 \, mol \, kg^{-1}$. $\Delta T_{b} = 2 	imes 0.512 	imes 0.1 = 0.1024 \, K$. The boiling point of pure water is $100^{\circ}C$. Therefore,the boiling point of the solution is $T_{b} = 100 + 0.1024 = 100.1024^{\circ}C$,which is closest to $100.10^{\circ}C$.

The boiling point (in ${}^{\circ}C$) of $0.1 \, molal$ aqueous solution of $CuSO_{4} \cdot 5H_{2}O$ at $1 \, bar$ is closest to $.....$
[Given : Ebullioscopic (molal boiling point elevation) constant of water,$K_{b}=0.512 \, K \, kg \, mol^{-1}$ ]

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The boiling point (in ${}^{\circ}C$) of $0.1 \, molal$ aqueous solution of $CuSO_{4} \cdot 5H_{2}O$ at $1 \, bar$ is closest to $.....$ [Given : Ebullioscopic (molal boiling point elevation) constant of water,$K_{b}=0.512 \, K \, kg \, mol^{-1}$ ]