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An aqueous solution of sugar undergoes acid catalysed hydrolysis. \(50 \mathrm{~g}\) sugar in \(125 \mathrm{~mL}\) water rotates the plane of plane polarized light by \(+13.1^{\circ}\) at \(\mathrm{t}=0 .\) After complete hydrolysis, it shows a rotation of \(-3.75^{\circ} .\) The percentage hydrolysis of sugar at time ' \(t\) ' in the same solution having a rotation of \(5^{\circ}\) is (a) \(42 \%\) (b) \(58 \%\) (c) \(48 \%\) (d) \(55 \%\)

Short Answer

Expert verified
The percentage hydrolysis of sugar at time 't' is 48% (choice c).

Step by step solution

01

Determine the Initial and Final Rotation

The initial rotation of the solution (c_{initial}) is given as \(+13.1^{\circ}\), and the final rotation (c_{final}) after complete hydrolysis is given as \(-3.75^{\circ}\). These values will be used to find the change in rotation.
02

Calculate the Total Change in Rotation Due to Hydrolysis

The total change in rotation (9C_{total}) due to complete hydrolysis is the difference between the initial and final rotations:\[\Delta C_{total} = C_{initial} - C_{final} = 13.1^{\circ} - (-3.75^{\circ}) = 16.85^{\circ}\]
03

Identify the Rotation Change at Time 't'

At time \(t\), the rotation of the solution is \(+5^{\circ}\). We need to calculate the change in rotation from the initial point to this time:\[\Delta C_t = C_{initial} - C_t = 13.1^{\circ} - 5^{\circ} = 8.1^{\circ}\]
04

Calculate Percentage Hydrolysis

The percentage hydrolysis of sugar can be calculated using the ratio of the change in rotation at time \(t\) to the total possible change, then multiplying by 100 to convert to a percentage:\[\text{Percentage hydrolysis} = \left(\frac{\Delta C_t}{\Delta C_{total}}\right) \times 100 = \left(\frac{8.1}{16.85}\right) \times 100 \approx 48\%\]
05

Select the Correct Answer Choice

Based on the calculated percentage hydrolysis of 48%, the correct choice for the answer is (c) \(48\%\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Optical Rotation
Understanding optical rotation is key to following this exercise. Optical rotation refers to how the plane of polarized light is twisted as it passes through a substance. This phenomenon occurs because certain molecules, like sugars, have the ability to rotate polarized light. In this case, an aqueous solution of sugar initially rotates light by \(+13.1^{\circ}\). This means that the solution turns the plane of light clockwise by this amount.
During hydrolysis—a chemical reaction where water breaks down a compound—the sugar changes, affecting optical rotation. The solution's optical rotation shifts from \(+13.1^{\circ}\) initially to \(-3.75^{\circ}\) after the reaction completes. This indicates a change in the solution's composition.
Monitoring this optical rotation over time gives insight into how much of the sugar is hydrolyzed at any moment. A decrease or change in the rotation value point towards the chemical transformation taking place through the reaction.
Percentage Hydrolysis
The percentage hydrolysis tells us the proportion of sugar in the solution that has reacted with water at a specific time. This is key for understanding how complete a reaction has become. To determine this, we need to compare the change in optical rotation at a given time with the total possible change.
The exercise starts with an initial optical rotation of \(+13.1^{\circ}\) and ends at \(-3.75^{\circ}\) after total hydrolysis. The solution gives a reading of \(+5^{\circ}\) at time 't'. To find the percentage hydrolysis, calculate how much rotation has occurred compared to the total potential change. This is the fraction \( \left( \frac{8.1}{16.85} \right) \) for our example, resulting in approximately 48%.
Calculating percentage hydrolysis helps verify if the reaction has proceeded as expected or if some conditions remain suboptimal.
Step-by-step Solution
Breaking down the exercise into manageable steps simplifies understanding complex reactions like hydrolysis.
  • **Step 1** focuses on identifying initial and final rotation values, \(+13.1^{\circ}\) and \(-3.75^{\circ}\) respectively. This frames the reaction's progress.
  • **Step 2** calculates the total shift by subtracting final from initial rotation values; \(13.1 - (-3.75) = 16.85\).
  • **Step 3** establishes current or interim rotation using the given timing's rotation \(+5^{\circ}\). Subtract this from the initial rotation: \(13.1 - 5 = 8.1\).
  • **Step 4** computes percentage using the relationship between the change at time \(t\) and complete hydrolysis converted to a percentage, yielding 48%.
  • Finally, **Step 5** involves matching the calculated percentage with available options, confirming 48% as the correct completion rate from the choices.
Simplifying the problem through these steps provides clarity, ensuring each calculation and logic follows the intended chemical pathway.
Aqueous Solution
An aqueous solution means the solvent used is water, which plays a special role in reactions like acid-catalyzed hydrolysis. The sugar dissolves in water; molecules are separated and surrounded by water molecules, an essential feature for reaction efficiency.
Water is not just a passive medium but actively participates in breaking down sugar molecules through hydrolysis. Acid catalysis further speeds up this process by effectively lowering the energy barrier required for the reaction. Hence, this solution setup allows for the controlled examination of reactions and observation of phenomena like optical rotation in a practical, real-time scenario.
Understanding the role of the aqueous environment helps explain why reactions can vary in rate and effectiveness under different conditions or solutions, advancing comprehension of kinetic and dynamic chemical concepts.

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Most popular questions from this chapter

The rate constant of a first-order reaction, \(\mathrm{A} \longrightarrow\) products, is \(60 \times 10^{-4} \mathrm{~s}^{-1} .\) Its rate at \([\mathrm{A}]=\) \(0.01 \mathrm{~mol} \mathrm{~L}^{-1}\) would be (a) \(60 \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}\) (b) \(36 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}\) (c) \(60 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}\) (d) \(36 \times 10^{-1} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}\)

For a certain reaction, the activation energy is zero. What is the value of rate constant at \(300 \mathrm{~K}\), if \(\mathrm{K}=1.6\) \(\times 10^{\circ} \mathrm{s}^{-1}\) at \(280 \mathrm{~K} ?\) (a) \(1.6 \times 10^{6} \mathrm{~s}^{-1}\) (b) zero (c) \(4.8 \times 10^{4} \mathrm{~s}^{-4}\) (d) \(3.2 \times 10^{12} \mathrm{~s}^{-1}\)

Which of the following statements is correct? (1) order of a reaction can be known from experimental results and not from the stoichiometry of a reaction. (2) molecularity a reaction refers to (i) each of the elementary steps in (an overall mechanism of) a complex reaction or (ii) a single step reaction. (3) overall molecularity of a reaction may be determined in a manner similar to overall order of reaction. (4) overall order of a reaction \(\mathrm{A}^{\mathrm{m}}+\mathrm{B}^{\mathrm{n}} \longrightarrow \mathrm{AB}_{\mathrm{x}}\) is \(\mathrm{m}+\mathrm{n}\) Select the correct answer using the following codes: (a) 2 and 3 (b) 1,3 and 4 (c) 2,3 and 4 (d) 1,2 and 3

The chemical kinetics of the reaction \(\mathrm{aA}+\mathrm{bB} \rightarrow\) \(\mathrm{C}\) at \(298 \mathrm{~K}\) were followed. The initial rates were recorded rates were recorded under different initial conditions and are summarized as follows. \begin{tabular}{lll} \hline Initial conc. \([\mathrm{A}]_{0}(\mathrm{~mol} / \mathrm{L})\) & Initial conc. \([\mathrm{B}]_{0}(\mathrm{~mol} / \mathbf{L})\) & Initial rate \((\mathrm{mol} / \mathrm{L} \mathrm{s})\) \\ \hline \(0.1\) & \(0.1\) & \(2.4 \times 10^{-3}\) \\ \(0.2\) & \(0.1\) & \(4.8 \times 10^{-3}\) \\ \(0.4\) & \(0.1\) & \(9.7 \times 10^{-3}\) \\ \(0.1\) & \(0.2\) & \(9.6 \times 10^{-3}\) \\ \(0.1\) & \(0.4\) & \(3.8 \times 10^{-2}\) \\ \hline \end{tabular} Which of the following statements is incorrect? (a) The rate constant \(\mathrm{k}\) is governed by the activation energy of the reaction (b) Reaction rate \(=\mathrm{k}[\mathrm{A}][\mathrm{B}]^{2}\) (c) In the chemical equation of \(a \mathrm{~A}+\mathrm{bB} \rightarrow \mathrm{C}, \mathrm{a}\) is 0 and \(b\) is 3 . (d) The overall order of reaction is third order.

If the half life period of a radioactive isotope is \(10 \mathrm{~s}\), then its average life will be (a) \(14.4 \mathrm{~s}\) (b) \(1.44 \mathrm{~s}\) (c) \(0.144 \mathrm{~s}\) (d) \(2.44 \mathrm{~s}\)

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