Question

How many liters of ethylene glycol antifreeze (C 2H 6O 2) would you add to your car radiator containing 15.0 L of water if you needed to protect your engine to –17.8°C? (The density of ethylene glycol is 1.1 g/mL. For water, K f = 1.86°C/m.)

Answer 1

`V=8.10L`

Step-by-step explanation:

Hello,

In this case, we can use the freezing point depression formula in order to compute the molality of the mentioned solute (ethylene glycol) considering a van't hoff factor of 1 since it is a covalent molecule:

`\Delta T=-i*m*Kf\\\\m=(\Delta T)/(i*Kf)=((-17.8-0)\°C)/(1*1.86\°C/m)=9.57m`

Next, since water density is 1 kg/L and molal units are mol/kg, we compute the present moles of solute:

`n=9.57mol/kg*15.0kg=143.55mol`

Next, the mass with its molar mass (62.07 g/mol):

`m=143.55mol*62.07g/mol=8910.05g`

Finally, with the given density we compute the required volume in liters:

`V=8910.05g*(1mL)/(1.1g) *(1L)/(1000mL)\\ \\V=8.10L`

Best regards.