When x times 10^{-2} mL methanol (molar mass | vedclass

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(C) The freezing point depression is given by $\Delta T_f = T_f^\circ - T_f = 273.15 \ K - 270.65 \ K = 2.5 \ K$.The formula for freezing point depres

Vedclass · Accepted Answer

$543$

Vedclass · Answer

(C) The freezing point depression is given by $\Delta T_f = T_f^\circ - T_f = 273.15 \ K - 270.65 \ K = 2.5 \ K$.The formula for freezing point depression is $\Delta T_f = K_f 	imes m$,where $m$ is the molality.$m = \frac{n_{methanol}}{w_{water} 	ext{ (in kg)}} = \frac{n}{0.1 \ kg}$.Substituting the values: $2.5 = 1.86 	imes \frac{n}{0.1} \Rightarrow n = \frac{2.5 	imes 0.1}{1.86} \approx 0.1344 \ mol$.The mass of methanol is $w = n 	imes M = 0.1344 \ mol 	imes 32 \ g \ mol^{-1} \approx 4.3008 \ g$.The volume of methanol is $V = \frac{w}{d} = \frac{4.3008 \ g}{0.792 \ g \ cm^{-3}} \approx 5.4303 \ mL$.Given $V = x 	imes 10^{-2} \ mL$,we have $5.4303 = x 	imes 10^{-2}$,so $x = 543.03$.Rounding to the nearest integer,$x = 543$.

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