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3 years ago

how many grams of antifreeze would be required per 500 g of water to prevent the water from feezing at a temperature of -39° C

Chemistry

1 answer:

andrezito [222]3 years ago

4 0

Answer:

333.7g of antifreeze

Explanation:

Freezing point depression in a solvent (In this case, water) occurs by the addition of a solute. The law is:

ΔT = Kf × m × i

Where:

ΔT is change in temperature (0°C - -20°C = 20°C)

Kf is freezing point depression constant (1.86°C / m)

m is molality of solution (moles solute / 0.5 kg solvent -500g water-)

i is Van't Hoff factor (1, assuming antifreeze is ethylene glycol -C₂H₄(OH)₂)

Replacing:

20°C = 1.86°C / m × moles solute / 0.5 kg solvent × 1

5.376 = moles solute

As molar mass of ethylene glycol is 62.07g/mol:

5.376 moles × (62.07g / 1mol) = <em>333.7g of antifreeze</em>.

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liberstina [14]

Answer:

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Explanation:

From the given information, since the molecular mass of the ion M+ is not given;

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2 years ago

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2 years ago

Calculate the molality of acetone in an aqueous solution with a mole fraction for acetone of 0.241. Answer in units of m.

Anit [1.1K]

Answer: The molality of solution is 17.6 mole/kg

Explanation:

Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.

$Molarity=\frac{n}{W_s}$

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n = moles of solute

$W_s$ = weight of solvent in kg

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3 years ago

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Answer:

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3 years ago

how many grams of antifreeze would be required per 500 g of water to prevent the water from feezing at a temperature of -39° C​

how many grams of antifreeze would be required per 500 g of water to prevent the water from feezing at a temperature of -39° C