The freezing point depression of 645 g of an | vedclass

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(A) $1$. Molar Mass Calculation:The molecular formula of ethylene glycol is $C_2H_6O_2$.Molar Mass $(M_2) = (2 	imes 12) + (6 	imes 1) + (2 	imes 1

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(A) $1$. Molar Mass Calculation:The molecular formula of ethylene glycol is $C_2H_6O_2$.Molar Mass $(M_2) = (2 	imes 12) + (6 	imes 1) + (2 	imes 16) = 62 \ g \ mol^{-1}$.$2$. Setup:Let the mass of ethylene glycol (solute) be $w_2 \ g$.Total mass of solution = $645 \ g$.Mass of water (solvent),$w_1 = (645 - w_2) \ g$.Given: $\Delta T_f = 2.25 \ K$ and $K_f = 1.86 \ K \ kg \ mol^{-1}$.$3$. Calculation:Using the freezing point depression formula: $\Delta T_f = \frac{K_f 	imes w_2 	imes 1000}{M_2 	imes w_1}$.Substituting the values: $2.25 = \frac{1.86 	imes w_2 	imes 1000}{62 	imes (645 - w_2)}$.$2.25 = \frac{30 	imes w_2}{645 - w_2}$.$2.25(645 - w_2) = 30w_2$.$1451.25 - 2.25w_2 = 30w_2$.$1451.25 = 32.25w_2$.$w_2 = \frac{1451.25}{32.25} = 45 \ g$.

The freezing point depression of $645 \ g$ of an aqueous solution of ethylene glycol $(C_2H_6O_2)$ is $2.25 \ K$. Find the weight of ethylene glycol in the solution. $[K_f = 1.86 \ K \ kg \ mol^{-1} ; H = 1, C = 12, O = 16 \ amu]$ (in $g$)

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