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Mekhanik [1.2K]
3 years ago
9

Winter is coming and it's time to make sure you have the right amount of antifreeze

Chemistry
1 answer:
ololo11 [35]3 years ago
5 0

The molality of the solution = 17.93 m

<h3>Further explanation</h3>

Given

6.00 L water with 6.00 L of  ethylene glycol(ρ=1.1132 g/cm³= 1.1132 kg/L)

Required

The molality

Solution

molality = mol of solute/ 1 kg solvent

mol of solute = mol of ethylene glycol

  • mass of ethylene glycol :

= volume x density

= 6 L x 1.1132 kg/L

= 6.6792 kg

= 6679.2 g

  • mol of ethylene glycol (MW=62.07 g/mol)

=mass : MW

=6679.2 : 62.07

=107.608

  • mass of water

6 L water = 6 kg water(ρ= 1 kg/L)

  • molality

\tt =\dfrac{107.608}{6}=17.93~m

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<h3>How to determine the mole of sodium sulphate Na₂SO₄</h3>
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<h3>How to determine the mass of sodium sulphate Na₂SO₄</h3>
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Mass = mole × molar mass

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Learn more about molarity:

brainly.com/question/15370276

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The Henry's law constant (kH) for O2 in water at 20°C is 1.28 × 10−3 mol/(L·atm). (a) How many grams of O2 will dissolve in 4.00
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