How many grams of methyl alcohol should be ad | vedclass

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(D) The depression in freezing point is given by $\Delta T_f = K_f 	imes m$,where $m$ is the molality of the solution.Given: $\Delta T_f = 273 \, K -

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$868.06$

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(D) The depression in freezing point is given by $\Delta T_f = K_f 	imes m$,where $m$ is the molality of the solution.Given: $\Delta T_f = 273 \, K - 268 \, K = 5 \, K$,$K_f = 1.86 \, K \, kg \, mol^{-1}$,volume of water = $10 \, L$,mass of water $(W)$ = $10 \, kg$,molar mass of methyl alcohol $(CH_3OH)$ = $32 \, g/mol$.Molality $m = \frac{w 	imes 1000}{M 	imes W_{solvent(g)}} = \frac{w}{32 	imes 10}$.Substituting the values: $5 = 1.86 	imes \frac{w}{32 	imes 10}$.$w = \frac{5 	imes 32 	imes 10}{1.86} = \frac{1600}{1.86} \approx 860.22 \, g$.Comparing with the given options,the closest value is $868.06 \, g$.

How many grams of methyl alcohol should be added to a $10 \, L$ tank of water to prevent its freezing at $268 \, K$? ($K_f$ for water is $1.86 \, K \, kg \, mol^{-1}$,density of water is $1 \, kg/L$)

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