Answer:
0.8342 grams of ethylene glycol must be added to one liter of water.
Explanation:
Volume of water or solvent = 1 L
Density of water = 1 g/L
Mass of the water = Density × Volume = 1 g/L × 1 L = 1 g
1 g = 0.001 kg
Molal freezing constant of water = 1.86 °C/m =1.86 °C/(mol/kg)
The van't Hoff factor contribution by ethylene glycol is 1 .
i = 1
[tex]\Delta T_f=25^oC[/tex]
[tex]\Delta T_f=i\times K_f\times m[/tex]
[tex]Molality=m(mol/kg)=\frac{\text{Moles of solute}}{\text{mass of solvent in kg}}[/tex]
[tex]25^oC=1\times 1.86 ^oC/(mol/kg)\times \frac{Moles}{0.001 kg}[/tex]
Moles of solute (ethylene glycol) = 0.013440860 moles
Mass of 0.013440860 moles of ethylene glycol:
[tex] 0.013440860 mol\times 62.07 g/mol=0.8342 g[/tex]
0.8342 grams of ethylene glycol must be added to one liter of water.
