2.7 kg of each of water and acetic acid are | vedclass

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(A) Moles of water $= \frac{2700 \ g}{18 \ g \ mol^{-1}} = 150 \ mol$.Moles of acetic acid $= \frac{2700 \ g}{60 \ g \ mol^{-1}} = 45 \ mol$.Since the

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$31$

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(A) Moles of water $= \frac{2700 \ g}{18 \ g \ mol^{-1}} = 150 \ mol$.Moles of acetic acid $= \frac{2700 \ g}{60 \ g \ mol^{-1}} = 45 \ mol$.Since the amount of water is greater than the amount of acetic acid,water acts as the solvent.The depression in freezing point is given by $\Delta T_f = K_f 	imes m$.Molality $(m) = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{45 \ mol}{2.7 \ kg} = 16.667 \ mol \ kg^{-1}$.$\Delta T_f = 1.86 \ K \ kg \ mol^{-1} 	imes 16.667 \ mol \ kg^{-1} \approx 31 \ K$.Freezing point of solution $= T_f^{\circ} - \Delta T_f = 0^{\circ} C - 31^{\circ} C = -31^{\circ} C$.Therefore,$x = 31$.

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