Skip to content
Intext Questions · 1.4

Q.Calculate the mass of urea (NH2CONH2NH_2CONH_2) required in making 2.5 kg of 0.25 molal aqueous solution.

CBSENCERTSubjective· 2mImportance★★★★★
13% · 17/131 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

To prepare a 0.25 molal aqueous urea solution, we must account for the mass of solvent, not the total solution. By setting up equations for molality and total mass, we find that 37.0 g of urea is required.

Molality is a measure of the concentration of a solute in a solution, defined as the number of moles of solute per kilogram of solvent. It is particularly useful because, unlike molarity, it does not change with temperature, as both moles and mass are temperature-independent quantities.

In this problem, we are given the molality of the solution and the total mass of the solution, not just the solvent. This is a common point of confusion. We need to determine the mass of urea (solute) required. To do this, we will set up equations based on the definition of molality and the total mass of the solution, then solve for the unknown mass of urea.

Here is the step-by-step calculation:

  1. Calculate the molar mass of urea (NH2CONH2NH_2CONH_2).

    The molar mass is the sum of the atomic masses of all atoms in the molecule.

    • Nitrogen (N): 2×14.01 g/mol=28.02 g/mol2 \times 14.01 \text{ g/mol} = 28.02 \text{ g/mol}
    • Hydrogen (H): 4×1.008 g/mol=4.032 g/mol4 \times 1.008 \text{ g/mol} = 4.032 \text{ g/mol}
    • Carbon (C): 1×12.01 g/mol=12.01 g/mol1 \times 12.01 \text{ g/mol} = 12.01 \text{ g/mol}
    • Oxygen (O): 1×16.00 g/mol=16.00 g/mol1 \times 16.00 \text{ g/mol} = 16.00 \text{ g/mol}

    Molar mass of urea (MureaM_{urea}) =28.02+4.032+12.01+16.00=60.062 g/mol= 28.02 + 4.032 + 12.01 + 16.00 = 60.062 \text{ g/mol}.

    We will use 60.06 g/mol60.06 \text{ g/mol} for our calculations.

  2. Define molality and set up initial relationships.

    Molality (mm) is given by:

    m=moles of solutemass of solvent (in kg)m = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}}

    Let nurean_{urea} be the moles of urea, MureaM_{urea} be the mass of urea (in grams), and MsolventM_{solvent} be the mass of solvent (water, in grams).

    From the definition of molality, we have:

0.25 mol/kg=nureaMsolvent/1000 g/kg0.25 \text{ mol/kg} = \frac{n_{urea}}{M_{solvent} / 1000 \text{ g/kg}}

The moles of urea can also be expressed as:

nurea=Murea (in g)60.06 g/moln_{urea} = \frac{M_{urea} \text{ (in g)}}{60.06 \text{ g/mol}}

Substituting the expression for $n_{urea}$ into the molality equation:

0.25=Murea/60.06Msolvent/10000.25 = \frac{M_{urea} / 60.06}{M_{solvent} / 1000}

0.25=Murea60.06×1000Msolvent0.25 = \frac{M_{urea}}{60.06} \times \frac{1000}{M_{solvent}}

0.25=1000Murea60.06Msolvent0.25 = \frac{1000 M_{urea}}{60.06 M_{solvent}}

Rearranging this equation to express $M_{solvent}$ in terms of $M_{urea}$:

0.25×60.06Msolvent=1000Murea0.25 \times 60.06 M_{solvent} = 1000 M_{urea}

15.015Msolvent=1000Murea15.015 M_{solvent} = 1000 M_{urea}

Msolvent=100015.015MureaM_{solvent} = \frac{1000}{15.015} M_{urea}

Msolvent≈66.606MureaM_{solvent} \approx 66.606 M_{urea}

  1. Use the total mass of the solution to solve for the mass of urea. …
Start your 14-day free trial to unlock the full solution →

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests
Start 14-day free trial →See Plans & Pricing →

Full access for 14 days. No credit card required.

←Previous question · 1.21.3 — Calculate the molarity of each of the following solutions: (a) 30 g of…→Next question · 1.21.5 — Calculate (a) molality (b) molarity and (c) mole fraction of KI if the…