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3 years ago

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An aqueous antifreeze solution is 60.0% ethylene glycol (HOCH2CH2OH) by mass and has a density of 1.06 g/mL. Calculate the molal

ity of ethylene glycol. Enter your answer to 1 decimal place.

1 answer:

galina1969 [7]3 years ago

4 0

Answer:

[HOCH₂CH₂OH] = 24.1 m

Explanation:

Ethylene glycol → HOCH₂CH₂OH

60% by mass means that 60 g of ethylene glycol are contained in 100 g of solution.

Solution mass = Solute mass + Solvent mass

100 g = 60 g + Solvent mass

Solvent mass = 40 g

Molality are the moles of solute contained in 1kg of solvent.

We determine the moles of solute → 60 g . 1mol/62 g = 0.967 moles

We convert the mass of solvent from g to kg → 40 g . 1kg/1000 g = 0.04 kg

Molality → 0.967 mol / 0.04 kg = 24.1 m

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Given that:

the temperature $T_1$ = 250 °C= ( 250+ 273.15 ) K = 523.15 K

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The truncated viral equation is expressed as:

$\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}$

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$T_c = 647.1 \ K \\ \\ P_c = 22055 \ kPa \\ \\ \omega = 0.345$

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$P__{\gamma}} = \frac{P}{P_c}$

$P__{\gamma}} = \frac{1800}{22055}$

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$B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}$

$B_o = 0.083 - \frac{0.422}{0.808^{1.6}}$

$B_o = 0.51$

$B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}$

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V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

Z = $\frac{PV}{RT}$

Z = $\frac{1800*10^3 *0.1249}{729.77*523.15}$

Z = 0.588

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