The molal elevation constant of water is 0.51 | vedclass

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(C) The elevation in boiling point is given by the formula: $\Delta T_b = K_b 	imes m$ Here,$K_b = 0.513 \ ^oC \ kg \ mol^{-1}$,Molality $(m)$ = $\fr

Vedclass · Accepted Answer

$100.26$

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(C) The elevation in boiling point is given by the formula: $\Delta T_b = K_b 	imes m$ Here,$K_b = 0.513 \ ^oC \ kg \ mol^{-1}$,Molality $(m)$ = $\frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{0.1 \ mol}{0.2 \ kg} = 0.5 \ mol \ kg^{-1}$ $\Delta T_b = 0.513 	imes 0.5 = 0.2565 \ ^oC$ Since the boiling point of pure water at $1 \ bar$ is $100 \ ^oC$,the boiling point of the solution is $T_b = T_b^0 + \Delta T_b = 100 + 0.2565 = 100.2565 \ ^oC \approx 100.26 \ ^oC$.

The molal elevation constant of water is $0.513 \ ^oC \ kg \ mol^{-1}$. If $0.1 \ mol$ of sugar is dissolved in $200 \ g$ of water,at what temperature $(^oC)$ will the solution start to boil under $1 \ bar$ pressure?

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