Question

How many grams of ethylene glycol (C2H6O2) must be added to 1.10 kg of water to produce a solution that freezes at -5.24 ∘C?

Answer #1

Ans :- **Grams of ethylene glycol required =**
**192.13 g**

**Explanation :-**

From Depression in freezing point , we have

ΔT_{f} = i x k_{f} x m
..................**(1)**

here, ΔT_{f} = Depression in freezing point =
T_{f}^{0} - T_{f}

T_{f}^{0} = freezing point of pure solvent =
0^{0}C and

T_{f} = freezing point of solution = - 5.24
^{0}C

k_{f} = Molal freezing point constant = 1.86°C
mol^{-1} kg for H_{2}O

i= Vant's haff factor = 1 (*for non-electrolyte like ethylene
glycol* )

m = Molality = Number of moles / Mass of solvent in kg

Number of moles of ethylene glycol = Mass (W) of ethylene glycol / Gram molar mass of ethylene glycol and

Gram molar mass of ethylene glycol = 62 g/mol

Substitute all these values in equation **(1)**, We
have

0 - (-5.24 ^{0}C) = 1 x 1.86°C mol^{-1} Kg x W /
62 g mol^{-1} x 1.10 Kg

5.24 = W x 0.027272727 g^{-1}

W = 5.24 / 0.027272727 g^{-1}

W = 5.24 / 0.027272727 g^{-1}

**W = 192.13 g**

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solution?
2.) What is the mass of 1.250 L of this solution (grams)?
3.) What is the concentration of this solution in ppm?
4.) What is the molality of the C2H6O2 solution?

An ethylene glycol solution contains 24.0 g of ethylene glycol
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Determine the freezing point of the solution.
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Determine the boiling point of the solution.
Express you answer in degrees Celsius.

An ethylene glycol solution contains 24.6 g of ethylene glycol
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