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zmey [24]
2 years ago
14

For each of the following choose the variable that will be more favorable in solution formation. A) Weak or Strong Solvent inter

molecular forces B) Weak or Strong Solute intermolecular forces C) Weak or Strong Solvent-Solute intermolecular force D) Increase or Decrease in Entropy E) Increase or Decrease in Enthalpy
Chemistry
1 answer:
Deffense [45]2 years ago
6 0

Based on factors affecting solution formation;

  • Strong solvent-solute intermolecular forces favors solution formation.
  • Increase in entropy favours solution formation
  • Decrease in enthalpy favours solution formation

<h3>What is a solution?</h3>

A solution is a substance formed when a substance known as solute dissolves in another substance know as solvent.

Factors that affect solution formation include:

  • strength of intermolecular forces between solute and solvent
  • entropy
  • enthalpy

Strong solvent-solute intermolecular forces favors solution formation.

Increase in entropy favours solution formation

Decrease in enthalpy favours solution formation.

Learn more about solutions at: brainly.com/question/6675586

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A student placed 15.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then
Ede4ka [16]

<u>Answer:</u> The mass of glucose in final solution is 1.085 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}      ......(1)

Given mass of glucose = 15.5 g

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Volume of solution = 100 mL

Putting values in equation 1, we get:

\text{Molarity of glucose solution}=\frac{15.5\times 1000}{180.2\times 100}\\\\\text{Molarity of glucose solution}=0.860M

To calculate the molarity of the diluted solution, we use the equation:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of the concentrated glucose solution

M_2\text{ and }V_2 are the molarity and volume of diluted glucose solution

We are given:

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Putting values in above equation, we get:

0.860\times 35.0=M_2\times 500\\\\M_2=\frac{0.860\times 35.0}{500}=0.0602M

Now, calculating the mass of glucose by using equation 1, we get:

Molarity of glucose solution = 0.0602 M

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

0.0602=\frac{\text{Mass of glucose solution}\times 1000}{180.2\times 100}\\\\\text{Mass of glucose solution}=\frac{0.0602\times 180.2\times 100}{1000}=1.085g

Hence, the mass of glucose in final solution is 1.085 grams

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