Question

How many grams of ethylene glycol (c2h6o2) must be added to 1.05 kg of water to produce a solution that freezes at -4.66 ∘c?

Answer 1

163g of ethylene glycol

Step-by-step explanation:

The addition of a non-volatile solute to a solvent as water produce the decreasing of freezing point following the formula:

ΔT = Kf×m×i

Where ΔT is change in freezing point (0°C- (-4.66°C) = 4.66°C); Kf is freezing point depression constant (1.86°C/m); m is molality (moles of solute / kg solvent); i is Van't Hoff factor (1 for ethylene glycol in water)

Replacing:

4.66°C = 1.86°C/m × moles solute / 1.05kg

2.63 = moles solute.

Molar mass of ethylene glycol is:

C: 12.01×2 = 24.02g/mol

H: 6×1.01 = 6.06g/mol

O: 2×16 = 32g/mol

24.02g/mol + 6.06g/mol + 32g/mol = 62.08g/mol

Thus, mass of ethylene glycol must be added is:

2.63mol × (62.08g/mol) = 163g of ethylene glycol

Answer 2

The formula for depression in freezing point is given by:

`\Delta T_(f)=m* K_(f)` (1)

where,

`\Delta T_(f)` = depression in freezing point

`K_(f)` = freezing point depression constant =
`1.86^(o)C kg/mol`

m = molality

First calculate the molality which is equal to

molality =
`(weight of solute in g)/(molecular weight of solute * weight of solvent in kg)`

=
`(w g)/(62 * 1.05 kg)`

=
`(w g)/(65.1 kg )`

Now,
`\Delta T_(f)` = freezing point of water - freezing point of solution

=
`0-(-4.66^(o)C)`

=
`4.66^(o)C`

Now, from formula (1)

`4.66^(o)C= (w g)/(65.1 kg)* 1.86^(o)C kg/mol`

w in g =
`(4.66* 65.1 )/(1.86)`

=
`163.1 g`

Thus, mass of ethylene glycol =
`163.1 g`