/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 4 If the solubility of glucose \(\... [FREE SOLUTION] | 91Ó°ÊÓ

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If the solubility of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) is \(120.3 \mathrm{~g} / 100 \mathrm{~mL}\) of \(\mathrm{H}_{2} \mathrm{O}\) at a given temperature, how many grams of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) can be dissolved in \(75.0 \mathrm{~mL}\) of \(\mathrm{H}_{2} \mathrm{O}\) ?

Short Answer

Expert verified
90.2 grams of glucose can be dissolved.

Step by step solution

01

Understand the Problem

We need to calculate the amount of glucose that can be dissolved in a given volume of water based on its solubility. The problem gives us the solubility as 120.3 g per 100 mL of water, and we need to find out how many grams will dissolve in 75.0 mL of water.
02

Set Up the Proportion

Using the given solubility, set up a proportion to find the mass of glucose that can be dissolved in 75.0 mL of water: \[ \frac{120.3 \text{ g}}{100 \text{ mL}} = \frac{x \text{ g}}{75.0 \text{ mL}} \]. This proportion will help us scale down the given solubility to the desired volume.
03

Solve the Proportion

Cross-multiply to solve for \(x\): \[ 120.3 \times 75.0 = 100 \times x \]. Calculate the left side: \[ 9022.5 = 100x \]. Divide both sides by 100 to find \(x\): \[ x = 90.225 \].
04

Round and Conclude

Round the calculated value of glucose to an appropriate number of significant figures, based on the given data. Here, we will round to three significant figures: \[ x = 90.2 \text{ g} \].Thus, 90.2 grams of glucose can be dissolved in 75.0 mL of water.

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Key Concepts

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

Glucose Dissolution
When glucose (a simple sugar with the chemical formula \(\mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}\)) is dissolved in water, it undergoes a process called dissolution. In dissolution, the glucose molecules become evenly distributed among the water molecules, resulting in a homogeneous solution. Water is often referred to as a 'universal solvent', due to its ability to dissolve a wide range of substances, including glucose.
This ability primarily stems from water molecules' polarity, allowing them to interact closely with glucose molecules, aiding the dissolution process.
The solubility of a substance like glucose indicates the maximum amount of it that can be fully dissolved in a specific amount of solvent, often expressed as grams per 100 mL of water. In the case at hand, glucose has a solubility of 120.3 grams per 100 mL. The term 'solubility' is crucial in chemistry to predict how substances will interact in a solution, providing guidance for mixing substances in practical applications.
Proportional Calculations
Proportional calculations are a fundamental mathematical concept used to determine how one quantity relates to another based on a constant ratio. In the context of solubility, these calculations help us find out how much of a solute (in this case, glucose) will dissolve in a different volume of solvent, given its known solubility at a standard volume.
We set up a proportion using the relationship:
  • Given solubility: \( \frac{120.3 \, \text{g}}{100 \, \text{mL}} \)
  • Desired amount of solvent: \(75.0 \, \text{mL}\)
The setup becomes a mathematical equation where we cross-multiply to find the unknown mass of glucose (\( x \)):
\[ \frac{120.3}{100} = \frac{x}{75.0} \]
This results in the equation:
\[ 120.3 \times 75.0 = 100 \times x \]

Upon solving, we determine the mass of glucose that can be dissolved in the specified arbitrary volume of water. This type of calculation is invaluable, allowing scientists and students alike to make quick and accurate predictions in chemistry based on straightforward arithmetic principles.
Significant Figures
Significant figures are the digits in a number that carry meaningful information regarding its precision. In the contexts of scientific measurements, it is crucial to employ significant figures as they convey the accuracy of calculated values based on the inputs used. While doing calculations, we often have to round figures to match the number of significant figures that are meaningful in the context of the problem statement.
For example, when calculating the mass of glucose that can dissolve, the solubility given (120.3 g/100 mL) and volume (75.0 mL) both have four significant figures. Thus, our final result (90.225 g) must be rounded to the same number of significant figures, resulting in 90.2 g.
  • This ensures the result retains the correct precision consistent with the initial data's precision.
  • Proper use of significant figures avoids overrepresentation or underrepresentation of the accuracy of a given measurement.
Understanding and applying the rule of significant figures is key to delivering precise and reliable scientific communication, especially when the derived results are relied upon for further calculations or practical applications.

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