Answer:
mass of ethylene glycol = 198. 562 g
Volume of ethylene glycol = 177.2875 ml
Explanation:
The freezing point depression of the solvent can be expressed by the formula:
[tex]\delta T = K_fmi[/tex]
where
[tex]K_f[/tex] = molar freezing point depression constant of water = 1.86 °C/m
[tex]\delta T[/tex] = freezing point depression = 0°C -(-29.75°C)
= 29.75°C
m = molarity of ethylene glycol
i = the value of Van't Hoff Factor = 1
replacing our value into above expression; we have:
[tex]29.75^0C= (1.86)*m*(1)[/tex]
[tex]m = \frac{29.75^0C} {(1.86)}[/tex]
m = 15.995 M
Also. Molarity = [tex]\frac{Mass(g) of solute}{Molar mass (g/mol) of solute}*\frac{1000}{mass of solvent(g)}[/tex]
[tex]15.995m = \frac{mass (g) of solute}{62.07g/mol C_2H_6O_2}*\frac{1000}{200}[/tex]
mass of ethylene glycol = 198. 562 g
Density = [tex]\frac{mass}{volume}[/tex]
Given that;
the density of ethylene glycol is 1.12g/ml
and our derived mass of ethylene glycol = 198. 562 g; we can determine how many milliliters of ethylene glycol are required by calculating the volume
Volume of ethylene glycol = [tex]\frac{mass}{density}[/tex]
Volume of ethylene glycol = [tex]\frac{198.562}{1.12}[/tex]
Volume of ethylene glycol = 177.2875 ml
