Problem 30 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

If \(250 \mathrm{~cm}^{3}\) of an aqueous solution containing \(0.73 \mathrm{~g}\) of a protein \(\mathrm{A}\) is isotonic with one litre of another aqueous solution containing \(1.65 \mathrm{~g}\) of a protein \(\mathrm{B}\), at \(298 \mathrm{~K}\), the ratio of the molecular masses of \(\mathrm{A}\) and \(\mathrm{B}\) is \( ____ \times 10^{-2}\) (to the nearest integer).

Step-by-Step Solution

Answer

The ratio is \(18 \times 10^{-2}\).

Step 1: Identify the Concept Used

The problem states that two solutions are isotonic. Isotonic solutions have equal osmotic pressures. The formula for osmotic pressure \( \Pi \) is:\[\Pi = \frac{n}{V}RT \]where \(n\) is the amount of substance in moles, \(V\) is the volume in liters, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin.

Step 2: Express Osmotic Pressure in Terms of Given Quantities

We know that \(n = \frac{mass}{molar\ mass} = \frac{m}{M}\). Therefore, osmotic pressure can be expressed as:\[\Pi = \frac{m}{MV}RT\]For isotonic solutions, \(\Pi_A = \Pi_B\). This means:\[\frac{m_A}{M_A V_A} RT = \frac{m_B}{M_B V_B} RT\]The \(RT\) terms cancel out, simplifying to:\[\frac{m_A}{M_A V_A} = \frac{m_B}{M_B V_B}\]

Step 3: Substitute Given Values

Plug the given values into the simplified equation. We have:- \(m_A = 0.73\, \text{g}\), \(V_A = 0.250\, \text{liters}\)- \(m_B = 1.65\, \text{g}\), \(V_B = 1.000\, \text{liters}\)Thus:\[\frac{0.73}{M_A \times 0.250} = \frac{1.65}{M_B \times 1.000}\]This simplifies to:\[\frac{0.73}{0.25M_A} = \frac{1.65}{M_B}\]

Step 4: Rearrange Equation to Find Ratio

Multiply both sides by \(M_A \cdot M_B\):\[0.73 \cdot M_B = 1.65 \cdot 0.25 \cdot M_A\]Rearrange the terms to find the ratio of \(M_A\) to \(M_B\):\[\frac{M_A}{M_B} = \frac{0.73}{1.65 \cdot 0.25}\]

Step 5: Calculate Numerical Value

Calculate the value:\[\frac{M_A}{M_B} = \frac{0.73}{0.4125} \approx 1.769\]Converting this to the form requested, \(1.769 \times 10^{0}\), equals approximately \(18\) when expressed as an integer in the form \( \times 10^{-2}\).

Step 6: Complete the Answer

Round 18 to the nearest integer and recognize it directly aligns with the requested format. As \(18 = 1800 \times 10^{-2}\), the ratio formats correctly as \( 18 \times 10^{-2} \).

Key Concepts

Osmotic Pressure

Osmotic pressure is a critical concept in understanding isotonic solutions. When two solutions are isotonic, they have equal osmotic pressures, meaning the flow of solvent across a semi-permeable membrane is balanced. Osmotic pressure (\(\Pi\)) can be calculated using the formula:\[\Pi = \frac{n}{V}RT\]Where:

• \(n\) is the number of moles of solute
• \(V\) is the volume of the solution in liters
• \(R\) is the ideal gas constant (\(0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1}\))
• \(T\) is the temperature in Kelvin

To find the osmotic pressure in terms of given mass and molar mass, use:\[\Pi = \frac{m}{MV}RT\]where \(m\) is the mass of solute and \(M\) is the molar mass. For two solutions to be isotonic,\(\Pi_A = \Pi_B\), implies that the ratios of their respective terms should be equal. Understanding this concept helps solve problems involving molecular mass when dealing with equal osmotic pressures across different solutions.

Molecular Mass

Molecular mass is an essential value for calculating osmotic pressure using solutions. It directly influences the number of moles of solute in a given mass, essentially connecting mass to molar concentration. The molecular mass (\(M\)) is defined as the mass of a molecule of a substance in atomic mass units (amu) or grams per mole (g/mol) in practical terms for chemistry problems.
When determining the molecular mass, it's crucial to use the correct information given in an isotonic problem as it can affect how you relate substances in a solution. In this context, the exercise demonstrates using the mass and volume of two isotonic solutions, equivalently compensating for different molecular masses to maintain the same osmotic pressure. By solving:\[\frac{m_A}{M_A V_A} = \frac{m_B}{M_B V_B}\]We can rearrange it to determine the ratio:\[\frac{M_A}{M_B} = \frac{m_A \cdot V_B}{m_B \cdot V_A}\]This equation perfectly reflects the inverse relationship between the mass of solute and its molecular mass in making up a solution's properties.

Chemical Calculations

Chemical calculations are at the heart of solving any chemistry-related problem involving isotonic solutions. These calculations often involve the direct use of formulas and rearrangement of expressions to find unknown variables based on given data, as demonstrated in our exercise. When working with isotonic solutions, including the molecular mass determination and using volume, the calculations aim to maintain the balance and equality of osmotic pressures. The process generally involves:

• Identifying the known versus unknown quantities
• Applying the correct formula, such as osmotic pressure calculation or moles related formulas
• Rearranging the formulas to express the unknown in terms of known variables
• Plugging in the numbers correctly to arrive at the solution

In the exercise provided, this involves working with the formula:\[\frac{m_A}{M_A V_A} = \frac{m_B}{M_B V_B}\]and carefully substituting given values of masses and volumes to find a ratio of molecular masses. It brings theory to practice by using systematic problem-solving methods in chemistry.