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Masja
2 months ago
3

156 g C12H22O11 (sucrose) is dissolved into 4.0 L of solution. What is the molar concentration of the solution? Report your answ

er with the correct number of significant figures.
Chemistry
1 answer:
Tems11 [2.7K]2 months ago
3 0

Answer:

The molar concentration calculates to 0.114 M (approximately 0.11 M)

Explanation:

Step 1: Provided data

Mass of sucrose = 156 grams

Molar mass of sucrose = 342.3 g/mol

Volume = 4.0 L

Step 2: Calculate moles of sucrose

Moles of sucrose = mass of sucrose / molar mass of sucrose

Moles of sucrose = 156 grams / 342.3 g/mol

This yields moles of sucrose = 0.456 moles

Step 3: Determine the concentration of the solution

Concentration of solution = moles of sucrose / volume

Concentration of solution = 0.456 moles / 4.0 L

The final concentration of the solution computes as 0.114 M

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Answer:

Complete Question:  

Equimolar quantities of CH3OH(l) and C2H5OH(l) are placed in separate 2.0 L containers that have been evacuated beforehand. Pressure gauges are attached to each container, and the temperature is maintained at 300 K. In both containers, liquid is consistently visible at the bottom. The varying pressure within the vessel that contains CH3OH(l) is illustrated below.

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Explanation:

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The London dispersion force is the least strong type of intermolecular force. It is a temporary force that arises when the electron arrangement in two neighboring atoms creates transient dipoles.  

The vapor pressure of a liquid reflects the equilibrium pressure of its vapor above the liquid (or solid); specifically, it represents the pressure associated with the evaporation of a liquid (or solid) in a sealed environment above the substance.

The pressure will be lower due to the stronger London dispersion forces acting between C2H5OH molecules compared to those between CH3OH molecules. This implies that when intermolecular forces are stronger, they intensify the interactions binding the substance together, thereby reducing the liquid's vapor pressure at any given temperature and making it more difficult to vaporize the substance.

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