Freezing Point Depression and Ethylene glycol

[sam.]

Homework Statement

Ethylene glycol, the primary ingredient in antifreeze, has the chemical formula C_2H_6O_2. The radiator fluid used in most cars is a half-and-half mixture of water and antifreeze.

What is the freezing point of radiator fluid that is 50% antifreeze by mass? K_f for water is 1.86 degrees Celsius/m.

Homework Equations

\DeltaT_f = K_f * m

m (molality) = # of moles of solute/mass of solution (kg)

The Attempt at a Solution

Okay, so I assumed 100 g of solution. And the molar mass of ethylene glycol is 62.08 g/mol. So you find the moles of ethylene glycol by multiplying 50 g by the molar mass which gives you 0.805 mol. Then you assume 100 g of water (which is 0.1 kg). And you solve for molality (m) and then you just sub it into the first equation? Is this correct?

 

[chemisttree]

Homework Statement

Ethylene glycol, the primary ingredient in antifreeze, has the chemical formula C_2H_6O_2. The radiator fluid used in most cars is a half-and-half mixture of water and antifreeze.

What is the freezing point of radiator fluid that is 50% antifreeze by mass? K_f for water is 1.86 degrees Celsius/m.

Homework Equations

\DeltaT_f = K_f * m

m (molality) = # of moles of solute/mass of solution (kg)

The Attempt at a Solution

Okay, so I assumed 100 g of solution. And the molar mass of ethylene glycol is 62.08 g/mol. So you find the moles of ethylene glycol by multiplying 50 g by the molar mass which gives you 0.805 mol. Then you assume 100 g of water (which is 0.1 kg). And you solve for molality (m) and then you just sub it into the first equation? Is this correct?

Not quite. Remember this:

m (molality) = # of moles of solute/mass of solution (kg)

do not "...assume 100 g of water (which is 0.1 kg)."

 

[Palestinian]

m (molality) = moles of solute/ mass of SOLVENT (KG) "NOTTTTTTTTTT SOLUTION"

 

[Borek]

You not only necroposted in the thread that is over two years old, but you also missed the fact that chemisttree already addressed the problem.

--

 

[chemisttree]

Boy! Those were the good old days. eh?