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Step-by-Step Solution
Step 1: Understand the concept of isotonic solutions
Two solutions are said to be isotonic when they have the same osmotic pressure
( \pi ). Osmotic pressure for any solution can be given by the formula
\pi = \dfrac{n}{V} RT , where n is the number of moles of the solute,
V is the volume (in liters), R is the gas constant, and T is the temperature in Kelvin.
Step 2: Set up the expression for osmotic pressure of Protein A
Let the molar mass of protein A be x g/mol. The mass of A dissolved is 0.73 g,
and volume of its solution is 250 cm3 = 0.25 L. Thus:
Number of moles of A = \dfrac{\text{mass of A}}{\text{molar mass of A}}
= \dfrac{0.73}{x} .
Therefore, osmotic pressure of A, \pi_A , will be
\pi_A = \left(\dfrac{0.73}{x \times 0.25}\right) RT .
Step 3: Set up the expression for osmotic pressure of Protein B
Let the molar mass of protein B be y g/mol. The mass of B is 1.65 g, and the volume
of its solution is 1 L. Thus:
Number of moles of B = \dfrac{\text{mass of B}}{\text{molar mass of B}}
= \dfrac{1.65}{y} .
Therefore, osmotic pressure of B, \pi_B , will be
\pi_B = \left(\dfrac{1.65}{y}\right) RT .
Step 4: Equate the osmotic pressures
Since the solutions are isotonic, \pi_A = \pi_B . Hence:
\dfrac{0.73}{x \times 0.25} \, RT = \dfrac{1.65}{y} \, RT .
RT cancels out on both sides, giving:
\dfrac{0.73}{0.25\,x} = \dfrac{1.65}{y} .
Step 5: Rearrange to find the ratio \dfrac{x}{y}
Rearranging the above equation:
\dfrac{x}{y} = \dfrac{0.73 / 0.25}{1.65}
= \dfrac{0.73}{0.25 \times 1.65} \approx 1.77 .
The problem asks for the ratio of the molecular masses of A to B (i.e. x/y )
in the form \_\_\_ \times 10^{-2} . Since 1.77 = 177 \times 10^{-2} ,
to the nearest integer it is
177 \times 10^{-2} .
