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3 years ago

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Now suppose, instead, that 5.678 g of a volatile solute is dissolved in 150.0 g of water. This solute also does not react with w

ater nor dissociate in solution. The pure solute displays, at 20°C, a vapour pressure of 1.754 torr. Again, assume an ideal solution. If, at 20°C the vapour pressure of this solution is also 17.344 torr. Calculate the molar mass of this volatile solute.

1 answer:

n200080 [17]3 years ago

4 0

Answer:

59.9 g/mol is the molar mass for the solute

Explanation:

Lowering vapor pressure → ΔP = P° . Xm

P° → Vapor pressure of pure solvent

ΔP = P° - Vapor pressure of solution

Xm = Mole fraction of solute

17.54 Torr - 17.344 Torr = 17.54 Torr . Xm

0.196 Torr / 17.54 Torr = Xm → 0.0112

These are the moles of solute / Total moles

Total moles = Moles of solute + Moles of solvent

We determine the moles of solvent → 150 g . 1mol/ 18 g = 8.33 moles

Now we can make this equation:

0.0112 = Moles of solute / Moles of solute + 8.33 mol

0.0112 Moles of solute + 0.0933 = Moles of solute

0.0933 = Moles of solute - 0.0112 Moles of solute

0.0933 = 0.9888 moles of solute → 0.0933 / 0.9888 = 0.0947 moles

Finally we can determine the molar mass (mol/g)

5.678 g / 0.0947 mol = 59.9 g/mol

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The following data were obtained from the question:

Original amount (N₀) = 0.15

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Next, we shall determine the amount remaining (N) after 5 half-life. This can be obtained as follow:

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We want to find out the n, so let's rearrange this:

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

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$0.02mol(\frac{62.03g}{mol})$

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