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lara [203]
3 years ago
7

An ethylene glycol solution contains 21.4 g of ethylene glycol (C2H6O2) in 97.6 mL of water. (Assume a density of 1.00 g/mL for

water.) Determine the freezing point and boiling point of the solution. (Assume a density of 1.00 g/ mL for water.)
Chemistry
1 answer:
8090 [49]3 years ago
4 0

Answer: The freezing point and boiling point of the solution are -6.6^0C and 101.8^0C respectively.

Explanation:

Depression in freezing point:

T_f^0-T^f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}

where,

T_f = freezing point of solution = ?

T^o_f = freezing point of water = 0^0C

k_f = freezing point constant of water = 1.86^0C/m

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

w_2 = mass of solute (ethylene glycol) = 21.4 g

w_1= mass of solvent (water) = density\times volume=1.00g/ml\times 97.6ml=97.6g

M_2 = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

(0-T_f)^0C=1\times (1.86^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}

T_f=-6.6^0C

Therefore,the freezing point of the solution is -6.6^0C

Elevation in boiling point :

T_b-T^b^0=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}

where,

T_b = boiling point of solution = ?

T^o_b = boiling point of water = 100^0C

k_b = boiling point constant of water = 0.52^0C/m

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

w_2 = mass of solute (ethylene glycol) = 21.4 g

w_1= mass of solvent (water) = density\times volume=1.00g/ml\times 97.6ml=97.6g

M_2 = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

(T_b-100)^0C=1\times (0.52^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}

T_b=101.8^0C

Thus the boiling point of the solution is 101.8^0C

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