Bond Burst

Category: Properties of Solution

  • The osmotic pressure of a solution containing 1.26 g of a protein dissolved in 200 mL of aqueous solution is 2.57×10−3 bar at 27°C. Determine the molar mass of the protein.

    A protein solution is prepared by dissolving 1.26 g of protein to make 200 mL of solution. At 27°C, the osmotic pressure of the solution is \(2.57\times10^{-3}\ \text{bar}\).

    Interpretation

    This question uses osmotic pressure to determine the molar mass of a protein. The underlying idea is quite elegant: although protein molecules are too large to count directly, their presence in solution creates an osmotic pressure that depends on the number of molecules present. By measuring that pressure, we can work backwards and determine the molar mass.

    For dilute solutions, osmotic pressure follows a relation analogous to the ideal gas equation:

    \(\pi = CRT\)

    where \(\pi\) is the osmotic pressure, \(C\) is the molar concentration, \(R\) is the gas constant, and \(T\) is the absolute temperature.

    Given Data

    Mass of protein:

    \(W = 1.26\ \text{g}\)

    Volume of solution:

    \(V = 200\ \text{mL}\)

    Temperature:

    \(T = 27^\circ\text{C} = 300\ \text{K}\)

    Osmotic pressure:

    \(\pi = 2.57\times10^{-3}\ \text{bar}\)

    Gas constant:

    \(R = 0.083\ \text{L·bar·mol}^{-1}\text{K}^{-1}\)

    Concept Application

    The osmotic pressure equation contains concentration, but the quantity we need is the molar mass.

    Since molar concentration is defined as

    \(C=\frac{\text{moles of solute}}{\text{volume of solution}}\)

    and the number of moles is

    \(n=\frac{W}{M}\)

    we can write

    \(C=\frac{W}{MV}\)

    Substituting this into the osmotic pressure equation gives

    \(\pi=\frac{WRT}{MV}\)

    This expression is especially useful because it directly connects the measurable quantities to the unknown molar mass.

    Solution

    Starting with

    \(\pi = CRT\)

    we obtain

    \(C=\frac{\pi}{RT}\)

    Substituting the given values:

    \(
    C=
    \frac{2.57\times10^{-3}}
    {0.083\times300}
    \) \(
    C=
    \frac{2.57\times10^{-3}}
    {24.9}
    \) \(
    C=1.032\times10^{-4}\ \text{mol L}^{-1}
    \)

    Now, expressing concentration in terms of mass and molar mass,

    \(
    C=\frac{W\times1000}{M\times V}
    \)

    Substituting the known values:

    \[
    \frac{1.26\times1000}{M\times200}
    = 1.0321285\times10^{-4}
    \]

    \[
    \frac{1260}{200M} = 1.0321285\times10^{-4}
    \]

    \(
    \frac{6.3}{M} = 1.0321285\times10^{-4}
    \)

    Solving for \(M\):

    \(
    M=
    \frac{6.3}{1.0321285\times10^{-4}}
    \) \(
    M=61038.9\ \text{g mol}^{-1}
    \)

    Therefore,

    \(
    M = 61039\ \text{g mol}^{-1}
    \)

    Answer

    \(
    Molar Mass = 61039\ \text{g mol}^{-1}
    \)

    Insight

    A powerful result worth remembering is

    \(
    \pi = \frac{WRT}{MV}
    \)

    This equation is simply the osmotic pressure equation rewritten in terms of the mass and molar mass of the solute. Whenever a problem provides mass of solute, volume of solution, temperature, and osmotic pressure, this relation should immediately come to mind.

    Notice the physical meaning of the answer as well. The calculated molar mass is about \(6.1\times10^4\ \text{g mol}^{-1}\), which is very large compared with ordinary molecules. That is exactly what we expect for a protein, reinforcing that the calculation is chemically reasonable.

  • In a mixture of solution,  50 mL of 16.9%(w/v) AgNO3 solution and 50mL  of 5.8%(w/v)NaCl solutions are mixed. Calculate the mass of the precipitate formed after mixing.

    1. 7 g
    2. 14 g
    3. 28 g
    4. 35 g

    Interpretation

    This is a precipitation reaction problem. When solutions of silver nitrate and sodium chloride are mixed, the ions exchange partners and form silver chloride, which is insoluble in water and therefore precipitates.

    The reaction is:AgNO3(aq)+NaCl(aq)→AgCl(s)+NaNO3(aq)\mathrm{AgNO_3(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_3(aq)}

    Notice that the stoichiometric ratio between AgNO3\mathrm{AgNO_3}​ and NaCl\mathrm{NaCl} is 1 : 1. Therefore, the amount of precipitate formed depends on whichever reactant is present in the smaller number of moles (the limiting reagent).

    Concept Application

    The concentrations are given in %(w/v).

    A x%x\%(w/v) solution means:x g solute in 100 mL solutionx\ \text{g solute in}\ 100\ \text{mL solution}

    So we can directly calculate the mass of each solute present in the given 50 mL portions.

    Solution

    Step 1: Calculate mass of AgNO3\mathrm{AgNO_3}​

    Given:16.9% (w/v) AgNO316.9\% \, (w/v)\ \mathrm{AgNO_3}

    This means:100 mL solution contains 16.9 g AgNO3100\ \text{mL solution contains }16.9\ \text{g AgNO}_3​

    Therefore, 5050 contains:16.9100×50=8.45 g\frac{16.9}{100}\times 50 =8.45\ \text{g}

    Step 2: Calculate moles of AgNO3\mathrm{AgNO_3}

    Molar mass of AgNO3\mathrm{AgNO_3}​:

    108+14+3(16)=170 g mol−1108+14+3(16)=170\ \text{g mol}^{-1}

    Hence,n(AgNO3)=8.45170=0.0497 moln(\mathrm{AgNO_3}) =\frac{8.45}{170} =0.0497\ \text{mol}

    Step 3: Calculate mass of NaCl\mathrm{NaCl}

    Given:5.8% (w/v) NaCl5.8\% \,(w/v)\ \mathrm{NaCl}

    Thus,100 mL solution contains 5.8 g NaCl100\ \text{mL solution contains }5.8\ \text{g NaCl}

    So 5050 mL contains:5.8100×50=2.9 g\frac{5.8}{100}\times 50 =2.9\ \text{g}

    Step 4: Calculate moles of NaCl\mathrm{NaCl}

    Molar mass of NaCl\mathrm{NaCl}:23+35.5=58.5 g mol−123+35.5=58.5\ \text{g mol}^{-1}

    Therefore,n(NaCl)=2.958.5=0.0496 moln(\mathrm{NaCl}) =\frac{2.9}{58.5} =0.0496\ \text{mol}

    Step 5: Identify the limiting reagent

    Reaction:AgNO3+NaCl→AgCl+NaNO3\mathrm{AgNO_3 + NaCl \rightarrow AgCl + NaNO_3}

    Required ratio:1:11:1

    Available moles:AgNO3=0.0497 mol\mathrm{AgNO_3}=0.0497\ \text{mol}NaCl=0.0496 mol\mathrm{NaCl}=0.0496\ \text{mol}

    Since NaCl\mathrm{NaCl} is slightly smaller, it is the limiting reagent.

    Hence,n(AgCl)=0.0496 moln(\mathrm{AgCl})=0.0496\ \text{mol}

    Step 6: Calculate mass of precipitated AgCl\mathrm{AgCl}

    Molar mass of AgCl\mathrm{AgCl}:108+35.5=143.5 g mol−1108+35.5=143.5\ \text{g mol}^{-1}

    Therefore,m(AgCl)=0.0496×143.5m(\mathrm{AgCl}) =0.0496\times 143.5=7.12 g=7.12\ \text{g}

    Answer

    Mass of AgCl precipitate formed ≈7.1 g\boxed{\text{Mass of AgCl precipitate formed } \approx 7.1\ \text{g}}

    Insight

    A useful shortcut for precipitation problems is:

    1. Convert %(w/v) directly into grams using the given volume.
    2. Convert grams to moles.
    3. Use the balanced equation to find the limiting reagent.
    4. Calculate the mass of the insoluble product.

    Also notice the elegant choice of concentrations here:16.9% AgNO3and5.8% NaCl16.9\% \text{ AgNO}_3 \quad\text{and}\quad 5.8\% \text{ NaCl}

    These give almost equal moles (≈0.05\approx 0.05 mol each), so nearly all of both reactants are consumed, producing about 0.050.05 mol of AgCl\mathrm{AgCl}. This makes the final answer easy to estimate even before doing the detailed calculation.

  • At what temperature does water boil at 101.325 kpa if 117g of NaCl is added to 222g of water in a saucepan?

    1. \(96.3^0C\)
    2. \(103.8^0C\)
    3. \(109.4^0C\)
    4. \(101.5^0C\)

    Interpretation

    This is a boiling point elevation problem, one of the important colligative properties of solutions.

    The key idea is that when a non-volatile solute is dissolved in a solvent, the vapor pressure of the solvent decreases. As a result, the liquid must be heated to a higher temperature before its vapor pressure becomes equal to the external pressure. Therefore, the boiling point increases.

    The governing relation is

    \(\Delta T_b=iK_bm\)

    where:

    • \(\Delta T_b\) = elevation in boiling point
    • \(i\) = van’t Hoff factor
    • \(K_b\) = boiling point elevation constant
    • \(m\) = molality of solution

    For sodium chloride:

    \(NaCl \rightarrow Na^+ + Cl^-\)

    One mole of NaCl produces two ions, so:

    \(i=2\)

    Concept Application

    Given:

    Mass of NaCl: \(117g\)

    Mass of water: \(222g=0.222kg\)

    Molar mass of NaCl: \(58.5gmol^{-1}\)

    Boiling point constant: \(K_b=0.52Kkgmol^{-1}\)

    Normal boiling point of water: \(100^\circ C\)

    We need to first find the molality, because \(K_b\) is used with molality.

    Solution

    Step 1: Calculate moles of NaCl

    \(
    \text{Moles of NaCl}=\frac{\text{mass of NaCl}}{\text{molar mass of NaCl}}
    \)

    Substituting:

    \(
    n=\frac{117}{58.5}
    \) \(
    n=2mol
    \)

    Step 2: Calculate molality

    Molality is:

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

    Therefore:

    \(
    m=\frac{2}{0.222}
    \) \(
    m=9.009molkg^{-1}
    \)

    Step 3: Calculate boiling point elevation

    Use:

    \(
    \Delta T_b=iK_bm
    \)

    Substitute the values:

    \(
    \Delta T_b=(2)(0.52)(9.009)
    \) \(
    \Delta T_b=9.37^\circ C
    \)

    So the boiling point increases by:

    \(
    9.37^\circ C
    \)

    Step 4: Calculate the new boiling point

    \(
    T_b=\text{normal boiling point}+\Delta T_b
    \) \(
    T_b=100+9.37
    \) \(
    T_b=109.37^\circ C
    \)

    Therefore:

    \(
    T_b = 109.4^\circ C
    \)

    Final Answer

    The chemically correct boiling point is:

    \(
    T_b = 109.4^\circ C
    \)