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tatuchka [14]
4 years ago
7

You have 1 liter of a freshly prepared 0.1 m sucrose (molecular mass 342) solution, which means ________.

Chemistry
2 answers:
iris [78.8K]4 years ago
8 0

1 liter of 0.1 M sucrose solution means that there is 34.2 g of sucrose dissolved in water to make a 1 liter solution. This also means that there are 6.022 × 10²² molecules of sucrose in this solution.  

<h3>FURTHER EXPLANATION </h3>

Molarity is a unit of concentration which describes the number of moles of solute dissolved per liter of a solution. It is calculated using the equation below:

molarity = \frac{moles \ of \ solute}{volume \ of \ solution}

Molarity is expressed with the unit mol/L or M (read as “molar”).  It is a very useful unit of concentration because many quantitative information about the solution can be obtained from this value.

<u>Mass of Solute from Molarity</u>

From the molarity of a solution, the equivalent mass of the solute used to prepare the solution may be obtained with the use of the formula mass (or molecular mass) through the equation:

molarity = \frac{\frac{mass\ of \ solute}{formula \ mass}}{volume \ of \ solution}\\\\mass \ of \ solute \ = molarity \times \ volume \ of \ solution \times formula \ mass

<u>Number of Representative Particles from Molarity</u>

Moreover, the number of molecules of solute present in the solution may also be obtained using the molarity and Avogadro's number.

molarity = \frac{moles \ of \ solute}{volume \ of \ solution}\\\\no. \ of \ representative \ particles \ = molarity \times volume \times 6.022 \times 10^{23} \ representative \ particles

For this problem, the mass of the solute dissolved in the solution and the number of sucrose molecules may be obtained from the molarity of the solution.

<em>Mass of Sucrose</em>

mass \ of \ sucrose \ = 0.1 \ \frac{mol}{L} \times 1 \ L \times \frac{342 \ g \ sucrose}{1 \ mol}\\\\\boxed {\boxed {mass \ of \ sucrose = 34.2 \ g \ sucrose}}

<em>Number of Sucrose Molecules</em>

<em>no. \ of \ sucrose \ molecules \ = 0.1 \frac{mol}{L} \times 1 \ L \times \frac{6.022 \times 10^{23} \ sucrose \ molecules}{1 \ mol}\\\\\boxed {\boxed {no. \ of \ sucrose \ molecules \ = 6.022 \times 10^{22} \ molecules}}</em>

<h3>LEARN MORE</h3>
  1. Ways of Expressing Concentration brainly.com/question/12936072
  2. Empirical Formula brainly.com/question/8516072

<em>Keywords: molarity, molar concentration</em>

Romashka-Z-Leto [24]4 years ago
8 0

One litre of a freshly prepared 0.1 M sucrose solution means that 34.2 g of sucrose is dissolved in water to get final 1 L solution.

Further Explanation:

Concentration

It is a term that is used to relatedifferent components of solution with each other. A variety of concentration terms are employed to attain this. Some of the concentration terms are mentioned below.

1. Molarity (M)

2. Mole fraction (X)

3. Molality (m)

4. Parts per million (ppm)

5. Mass percent ((w/w) %)

6. Volume percent ((v/v) %)

7. Parts per billion (ppb)

Molarity is defined as moles of solute present in one litre of solution. It is represented by M and its unit is mol/L. The expression for molarity of solution is as follows:

{\text{Molarity of solution}} = \dfrac{{{\text{Moles }}\left( {{\text{mol}}} \right){\text{of solute}}}}{{{\text{Volume }}\left( {\text{L}} \right){\text{ of solution}}}}

We are provided with 0.1 M sucrose solution. According to the definition of molarity, this indicates that 0.1 moles of sucrose are dissolved in 1 L of the solution.  

The formula to calculate the moles of sucrose is as follows:

{\text{Moles of sucrose}} = \dfrac{{{\text{Mass of sucrose}}}}{{{\text{Molar mass of sucrose}}}}                                            …… (1)

Rearrange equation (1) for mass of sucrose.

{\text{Mass of sucrose}} = \left( {{\text{Moles of sucrose}}} \right)\left( {{\text{Molar mass of sucrose}}} \right)                  …… (2)

Substitute 0.1 mol for moles of sucrose and 342 g/mol for molar mass of sucrose in equation (2).

\begin{aligned}{\text{Mass of sucrose}} &= \left( {{\text{0}}{\text{.1 mol}}} \right)\left( {{\text{342 g/mol}}} \right) \\&= {\text{34}}{\text{.2 g}} \\\end{aligned}  

Therefore 34.2 g of sucrose is dissolved in water to get final 1 L solution.

Learn more:

  1. Calculation of volume of gas: brainly.com/question/3636135
  2. Determine how many moles of water produce: brainly.com/question/1405182

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Concentration terms

Keywords: concentration, concentration terms, solutions, molarity, molality, sucrose, moles, mass, molar mass, 342 g/mol, 34.2 g, 0.1 M, mass of sucrose.

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<h3>Answer:</h3>

5.55 mol C₂H₅OH

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Tables
  • Moles

<u>Stoichiometry</u>

  • Using Dimensional Analysis
  • Analyzing Reactions RxN
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

[Given] 500. g C₆H₁₂O₆ (Glucose)

[Solve] moles C₂H₅OH (Ethanol)

<u>Step 2: Identify Conversions</u>

[RxN] 1 mol C₆H₁₂O₆ → 2 mol C₂H₅OH

[PT] Molar mass of C - 12.01 g/mol

[PT] Molar Mass of H - 1.01 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of C₆H₁₂O₆ - 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol

<u>Step 3: Stoichiometry</u>

  1. [DA] Set up conversion:                                                                                 \displaystyle 500 \ g \ C_6H_{12}O_6(\frac{1 \ mol \ C_6H_{12}O_6}{180.18 \ g \ C_6H_{12}O_6})(\frac{2 \ mol \ C_2H_5OH}{1 \ mol \ C_6H_{12}O_6})
  2. [DA} Multiply/Divide [Cancel out units]:                                                         \displaystyle 5.55001 \ mol \ C_2H_5OH

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

5.55001 mol C₂H₅OH ≈ 5.55 mol C₂H₅OH

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