The molal elevation constant of water is 0.51 | vedclass

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(B) The elevation in boiling point is given by the formula: $\Delta T_b = i 	imes K_b 	imes m$. For $NaCl$,the van't Hoff factor $i = 2$ (since $NaC

Vedclass · Accepted Answer

$100.1$

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(B) The elevation in boiling point is given by the formula: $\Delta T_b = i 	imes K_b 	imes m$. For $NaCl$,the van't Hoff factor $i = 2$ (since $NaCl$ dissociates into $Na^+$ and $Cl^-$). Given $K_b = 0.51 \ K \ kg \ mol^{-1}$ and $m = 0.1 \ m$. $\Delta T_b = 2 	imes 0.51 	imes 0.1 = 0.102 \ ^oC$. The boiling point of the solution is $T_b = T_b^0 + \Delta T_b$. Since the boiling point of pure water $T_b^0 = 100 \ ^oC$,the boiling point of the solution is $100 + 0.102 = 100.102 \ ^oC$,which is approximately $100.1 \ ^oC$.

The molal elevation constant of water is $0.51 \ K \ kg \ mol^{-1}$. The boiling point of a $0.1 \ m$ aqueous $NaCl$ solution is approximately ......... $^oC$.

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