Respuesta :
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
Explanation:
Depression in freezing point is given by:
[tex]\Delta T_f=i\times K_f\times m[/tex]
[tex]\Delta T_f=T_f^0-T_f=(0-(-10))^0C=10^0C[/tex] = Depression in freezing point
i= vant hoff factor ( for non electrolytes , i= 1)
[tex]K_f[/tex]= freezing point constant for water= [tex]1.86^0C/m[/tex]
m= molality
[tex]\Delta T_f=i\times K_f\times \frac{\text{mass of solute}\times 1000}{\text{molar mass of solute}\times \text{weight of solvent in g}}[/tex]
weight of solvent (water ) = [tex]density\times Volume[/tex]
weight of solvent (water) =[tex]1.0g/ml\times 10000ml=10000g[/tex] ( 1L=1000ml)
[tex]10^0C=1\times 1.86^0C/m\times \frac{x\times 1000}{46\times 10000g}[/tex]
[tex]x=2473g[/tex]
Thus 2473 g of ethanol must be added to 10.0 L of water to give a solution that freezes at −10.0°C
