Chapter 12: Problem 54

How many grams of urea \(\left[\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}\right]\) must be added to \(450 \mathrm{~g}\) of water to give a solution with a vapor pressure \(2.50 \mathrm{mmHg}\) less than that of pure water at \(30^{\circ} \mathrm{C}\) ? (The vapor pressure of water at \(30^{\circ} \mathrm{C}\) is \(31.8 \mathrm{mmHg} .\) )

Short Answer

Expert verified

The amount of urea that must be added to 450 g of water to give a solution with a vapor pressure 2.50 mmHg less than that of pure water at 30°C is approximately 0.173 g.

Step by step solution

- Formula & Data

We first write down Raoult's law in the following form: ΔP = Kb * m, where ΔP is the decrease in vapor pressure, Kb is the molal boiling point elevation constant for water (which is 0.512 °C.kg/mol), and m is the molality. Given data: ΔP= 2.50mmHg, we will convert ΔP into atmospheres because Kb is given in terms of atmospheres. Also, we are given that the mass of water (solvent) = 450 g, which is 0.450 kg. Finally, we also need to know the molar mass of urea, which is 60 g/mol.

- Conversion to proper units

First, the change in pressure (ΔP) should be converted from mmHg to atmospheres, as the Kb value is given in terms of atmospheres. 1 atm is approximately equal to 760 mmHg, so the ΔP in atmospheres is ΔP(atm) = 2.5mmHg / 760 mmHg/atm = 0.0032895 atm. Use this value for future calculations.

- Calculate Molality

Using the Raoult's law, we can solve for the molality. On substituting the values of ΔP(atm) = 0.0032895 atm and Kb = 0.512 °C.kg/mol into the equation, we can solve for molality (m) which comes out to be m = ΔP / Kb = 0.0032895 / 0.512 = 0.00642 mol/kg.

- Calculate Moles of Urea

Now, we will use the formula of molality (moles of solute / kg of solvent) to calculate the moles of urea (NH2)2CO. Moles of Urea = Molality * kg of water = 0.00642 * 0.450 = 0.002889 moles.

- Calculate Grams of Urea

Now that we have the moles of urea, we convert it into grams using the molar mass. Mass of Urea = Moles * Molar mass = 0.002889 moles * 60 g/mol = 0.173 g.

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Short Answer

Step by step solution

- Formula & Data

- Conversion to proper units

- Calculate Molality

- Calculate Moles of Urea

- Calculate Grams of Urea

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