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Untitled - Kelly Walsh High School

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Solutions 175<br />

Don’t Forget!<br />

Don’t Forget!<br />

In preparing a molar solution, the correct number of moles of solute (commonly<br />

converted to grams using the molar mass) is dissolved and diluted to the<br />

required volume.<br />

Molality<br />

Molality (m) is the moles of solute per kilogram of solvent.<br />

m = moles solute/kg solvent<br />

Notice that it is kilograms of solvent, not solution. In the other concentration<br />

units, it has been the mass or volume of the solution. In molal solutions, it is the<br />

mass of the solvent.<br />

Ethylene glycol (C 2H 6O 2) is in antifreeze. Determine the molality of ethylene glycol<br />

in a solution prepared by adding 31.0 g of ethylene glycol to 200.0 g of water.<br />

Molality a 31.0 g C2H6O2 g<br />

ba1000<br />

200.0 g H2O 1kg ba1 mol C2H6O2 b 2.50 m C2H6O2 62.1 g C2H6O2 It is the kilograms of the solvent in the denominator and not the solution.<br />

Mole Fraction<br />

The mole fraction, X, is the moles of solute divided by the total number of<br />

moles of solution.<br />

X moles solute/total moles of solution<br />

Suppose you have a solution containing 0.50 mol of sodium chloride in 2.0 mol<br />

of water. The mole fraction of sodium chloride would be:<br />

XNaCl moles NaCl/(moles NaCl moles H2O) 0.50/(0.50 2.00) 0.20

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