Problem 39 - A Complete Resource Book in Chemistry for JEE Main

Question

A \(5 \%\) solution of sugarcane \((\) Mol \(w t=342)\) is isotonic with \(1 \%\) solution of \(X\) under similar conditions. The mol. Wt of \(X\) is: (a) \(136.2\) (b) \(68.4\) (c) \(34.2\) (d) \(171.2\)

Step-by-Step Solution

Answer

The molecular weight of X is 68.4 (Option b).

Step 1: Understand the concept of isotonic solutions

Isotonic solutions have the same osmotic pressure. Therefore, the number of solute particles (in moles) per unit volume must be equal for both solutions.

Step 2: Calculate moles of sugarcane in the solution

A 5% (w/v) solution of sugarcane means 5 grams of sugarcane per 100 mL of solution. The moles of sugarcane can be calculated using its molar mass. \[ \text{Moles of sugarcane} = \frac{5}{342} \text{ moles per } 0.1 \text{ L} \]

Step 3: Calculate the molarity of sugarcane solution

Molarity is the number of moles per liter. Using the number of moles calculated in Step 2, convert them to molarity: \[ \text{Molarity of sugarcane solution} = \frac{5}{342 \times 0.1} \text{ M} \]

Step 4: Express 1% w/v solution of X in terms of molarity

A 1% (w/v) solution of X means 1 gram of X per 100 mL of solution, which is equal to 0.1 L. If the molecular weight of X is \(M_X\), then:\[ \text{Molarity of } X = \frac{1}{M_X \times 0.1} \text{ M} \]

Step 5: Equate the molarities of sugarcane and X

Since the solutions are isotonic, their molarities are equal:\[ \frac{5}{342 \times 0.1} = \frac{1}{M_X \times 0.1} \]

Step 6: Solve for the molecular weight of X

Solve the equation from Step 5 to find the molecular weight:\[ \frac{5}{342} = \frac{1}{M_X} \quad \Rightarrow \quad M_X = 342 \times \frac{1}{5} = 68.4 \]

Key Concepts

Osmotic Pressure

When two solutions are isotonic, they have equal osmotic pressure.
Osmotic pressure is a key concept in chemistry, describing the tendency of a solvent to move through a semipermeable membrane.
This movement is from a region of lower solute concentration to a region of higher solute concentration.

• To put it simply, osmotic pressure causes water to "balance out" between different solutions.
• This is crucial for maintaining equilibrium between the solutions.
• In our exercise, both the 5% sugarcane solution and the 1% solution of compound X must exert the same osmotic pressure to be isotonic.

Understanding isotonicity helps in biology and medical fields, particularly when administering intravenous fluids.
They need to match the body’s osmotic pressure to avoid disrupting cell integrity.

Molarity

Molarity is a measure of the concentration of a solute in a solution. It’s expressed as the number of moles of solute per liter of solution.

• This concept helps us understand how concentrated a particular solution is.
• In our problem, the sugarcane solution has a molarity calculated using its known weight and molecular mass.

The equation used for calculating molarity is:\[\text{Molarity (M)} = \frac{\text{Moles of Solute}}{\text{Liters of Solution}}\]In the solution, we first calculated moles of sugarcane per 100 mL (or 0.1 L) from the weight percentage.
That way, we can understand how much solute is present per unit volume, a crucial step in comparing the two isotonic solutions.
Thus, making calculations with molarity aligns with achieving the understanding of the solution's concentration, which is directly linked to its osmotic pressure.

Molecular Weight Calculation

Finding the molecular weight of a compound is essential, especially when dealing with isotonic solutions.

• The molecular weight tells us how much one mole of the substance weighs in grams.
• To solve the problem, the molecular weight of substance X was calculated using the molar equivalence of sugarcane as both solutions are isotonic.

The calculation follows from the isotonic condition:\[\frac{5}{342 \times 0.1} = \frac{1}{M_X \times 0.1}\]By solving this equation, the molecular weight of X \( (M_X) \) was found to be 68.4 g/mol.
Understanding molecular weights helps in stoichiometric calculations and other chemical analyses where precise measurements are required.
This approach not only aids in solving the exercise but helps in exploring how different concentrations interact in real-life applications, such as creating solutions with desired osmotic pressures.