Question

What mass of ethanol, C2H5OH a nonelectrolyte, must be added to 10.0 L of water to give a solution that freezes at −10.0°C? Assume the density of water is 1.0 g/mL. Kf of water is 1.86°C/m.

Answer 1

Answer: Thus 2473 g of ethanol must be added to 10.0 L of water to give a solution that freezes at −10.0°C

Step-by-step explanation:

Depression in freezing point is given by:

`\Delta T_f=i* K_f* m`

`\Delta T_f=T_f^0-T_f=(0-(-10))^0C=10^0C` = Depression in freezing point

i= vant hoff factor ( for non electrolytes , i= 1)

`K_f`= freezing point constant for water=
`1.86^0C/m`

m= molality

`\Delta T_f=i* K_f* \frac{\text{mass of solute}* 1000}{\text{molar mass of solute}* \text{weight of solvent in g}}`

weight of solvent (water ) =
`density* Volume`

weight of solvent (water) =
`1.0g/ml* 10000ml=10000g` ( 1L=1000ml)

`10^0C=1* 1.86^0C/m* (x* 1000)/(46* 10000g)`

`x=2473g`

Thus 2473 g of ethanol must be added to 10.0 L of water to give a solution that freezes at −10.0°C