The freezing point of a solution containing 1 | vedclass

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(B) Given: $(T_f)_s = 271.9 \ K$,$w = 1.25 \ g$,$W = 20 \ g$,$K_f = 1.86 \ K \ kg \ mol^{-1}$,$T_0 = 273 \ K$. $\Delta T_f = T_0 - (T_f)_s = 273 - 271

Vedclass · Accepted Answer

$105.68$

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(B) Given: $(T_f)_s = 271.9 \ K$,$w = 1.25 \ g$,$W = 20 \ g$,$K_f = 1.86 \ K \ kg \ mol^{-1}$,$T_0 = 273 \ K$. $\Delta T_f = T_0 - (T_f)_s = 273 - 271.9 = 1.1 \ K$. The formula for depression in freezing point is: $\Delta T_f = \frac{w 	imes 1000 	imes K_f}{M_w 	imes W}$. Rearranging for molar mass $(M_w)$: $M_w = \frac{w 	imes 1000 	imes K_f}{\Delta T_f 	imes W}$. Substituting the values: $M_w = \frac{1.25 	imes 1000 	imes 1.86}{1.1 	imes 20}$. $M_w = \frac{2325}{22} = 105.68 \ g \ mol^{-1}$.

The freezing point of a solution containing $1.25 \ g$ of a non-electrolyte solute in $20 \ g$ of water is $271.9 \ K$. What is the molar mass of the solute? (Given: $K_f$ for water = $1.86 \ K \ kg \ mol^{-1}$,Freezing point of pure water = $273 \ K$)

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