Ethylene glycol is termed as the primary ingredients in antifreeze. The ethylene glycol molecular formula is Cu2082Hu2086Ou2082. Molar mass of Cu2082Hu2086Ou2082 is = (2×12) +(6×1) + (216) = 62g/mol Now that antifreeze by mass is 50%, then there is 1kg of ethylene glycol which is present in 1kg of water. u0394Tf = Kfxm u0394Tf = depression in the freezing point. = freezing point of water freezing point of the solution = Ou00b0c – Tf = -Tf Kf = depression in freezing constant of water = 1.86u00b0C/m M is the molarity of the solution. =(mass/molar mass) mass of solvent in kg =1000g/62 (g/mol) /1kg =16.13m If we plug the value we get -Tf = 1.86x 16.13 = 30 Tf = -30u00b0c

## Answers ( 2 )

Answer:-30.02 u00baCExplanation:Assuming the antifreeze to be ethylene glycol (Cu2082Hu2086Ou2082) which is popular antifreeze.

Molar mass of ethylene glycol (Cu2082Hu2086Ou2082) = 62g/mol

calculate the freezing point depression of the solutionStep 1:u0394T = -Kf*Mwhere,u0394T= depression in the freezing point.

M = the molarity of the solution (mol solute / Kg solvent)

Kf = molar freezing point constant of water = 1.86u00b0C/m

To determine depression in the freezing point (u0394T), first we need to calculate;calculate the molarity of the solute (ethylene glycol)Step 2:Molar mass ethylene glycol = 62 g/mol

molarity of ethylene glycol in mol = 50 g / 62g/mol =

0.807 molcalculate mass of solvent in kgStep 3:There is 1kg of ethylene glycol which is present in 1kg of water

mass of solvent (water) in kg= 50 g/ 1000 g/ Kg =

0.050 Kgcalculate the molarity of the solution (M)Step 4:M = 0.807 mol / 0.050 Kg =

16.14 mStep 5:u0394T = – Kf*M = -1.86 u00baC/m x 16.14 m

= -30.02 u00baCEthylene glycol is termed as the primary ingredients in antifreeze.

The ethylene glycol molecular formula is Cu2082Hu2086Ou2082.

Molar mass of Cu2082Hu2086Ou2082 is = (2×12) +(6×1) + (216) = 62g/mol

Now that antifreeze by mass is 50%, then there is 1kg of ethylene glycol which is present in 1kg of water.

u0394Tf = Kfxm

u0394Tf = depression in the freezing point.

= freezing point of water freezing point of the solution

= Ou00b0c – Tf

= -Tf

Kf = depression in freezing constant of water = 1.86u00b0C/m

M is the molarity of the solution.

=(mass/molar mass) mass of solvent in kg

=1000g/62 (g/mol) /1kg

=16.13m

If we plug the value we get

-Tf = 1.86x 16.13 = 30

Tf = -30u00b0c