Question

A \(5 \%\) (w/V) solution of cane sugar (molecular weight \(=342\) ) is isotonic with \(1 \%\) (w/V) solution of a substance \(X .\) The molecular weight of \(X\) is: (a) \(34.2\) (b) \(171.2\) (c) \(68.4\) (d) \(136.8\)

Short answer

The molecular weight of X is 34.2

Step-by-step solution

Understanding Isotonic Solutions

Isotonic solutions have the same osmotic pressure, which means they have the same number of particles, or moles, of solute per liter of solution. Since the 5% cane sugar solution is isotonic with the 1% solution of substance X, they must have the same molar concentration.

Calculating the Molarity of the Cane Sugar Solution

To calculate molarity of the cane sugar solution, we use the formula: Molarity (M) = (Percent by weight (w/v) * 10) / Molecular weight (MW). For a 5% w/v cane sugar solution: M = (5 * 10) / 342. Calculate the molarity using the given information.

Calculating the Molarity of Substance X Solution

Now, the molarity of the substance X solution must be the same as that of the cane sugar solution since they are isotonic. For a 1% w/v solution of X, if MW is its molecular weight, then M = (1 * 10) / MW. This molarity should be equal to that of the cane sugar solution calculated in the previous step.

Determining the Molecular Weight of Substance X

Set the molarities equal to each other and solve for MW, the molecular weight of substance X. From previous steps, we have (5 * 10) / 342 = (1 * 10) / MW. Simplify and solve for MW.

Key concepts

Osmotic Pressure in Isotonic Solutions

When studying solutions in biology or chemistry, it's essential to understand the concept of osmotic pressure, which is critical to the phenomenon of isotonicity. Osmotic pressure is defined as the pressure required to stop the flow of water through a semipermeable membrane separating two solutions with different concentrations. In the case of isotonic solutions, this pressure is balanced because the concentration of dissolved particles (usually measured in moles) is the same on both sides of the membrane. This balance means that there is no net movement of water in or out of cells, which is vital for maintaining cell integrity and function.

In our exercise, we use the principle that two isotonic solutions have the same osmotic pressure to determine an unknown molecular weight. This is because isotonic solutions have equal molar concentrations, a property we leverage to calculate the unknown. The beauty of isotonic solutions lies in their application – from medical treatments like IV fluids to everyday scenarios where cell homeostasis is imperative.

Molarity Calculation Explained

The next pivotal concept in our exploration is molarity calculation. Molarity, often denoted as 'M,' is a measure of concentration that expresses the moles of solute per liter of solution. It serves as a uniform way to compare the strength of solutions and is a cornerstone principle in chemistry and biology when dealing with reactions and solute-solvent interactions.

To calculate the molarity, the formula is simple: \[ \text{Molarity (M)} = \frac{\text{Percent by weight (w/v)} \times 10}{\text{Molecular weight (MW)}} \]
This equation plays a crucial role in our textbook problem — by knowing the percent concentration by weight and the molecular weight, we can easily find the molarity of the cane sugar solution. Then, using the concept of isotonic solutions, we assume that the unknown solution of substance X has the same molarity. This assumption is the key to unlocking the molecular weight of X.

Determining Molecular Weight from Isotonic Solutions

Finally, we delve into the third core concept, ‘molecular weight’. The molecular weight (also called molecular mass) is the weight of one mole of a substance and is measured in grams per mole (g/mol). It is a sum of the atomic weights of all atoms in the molecule. Knowing the molecular weight is essential for converting between moles of a substance and its weight in grams — foundational knowledge for any kind of chemical calculation involving elements or compounds.

Considering our exercise, we utilized the molarities of two isotonic solutions to determine the unknown molecular weight of the substance X. The relationship between percent concentration and molecular weight allowed us to establish a linear equation, from which the molecular weight of X can be ascertained. By understanding these interconnected concepts, students can apply their knowledge to a broad range of real-world and theoretical situations involving chemical solutions and biological fluids.