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What is the molarity of a solution that is made by dissolving 3.68 g of sucrose (c12h22o11) in sufficient water to form 275.0 ml of solution?

Respuesta :

meerkat18
In dealing with mixtures in chemistry, the concentration of the solute in the solution may be expressed in different ways. It could be in terms of molarity, molality, normality, percentages, parts per million, mass per volume, and many more. The most common unit of concentration is molarity which is in terms of moles solute per liter solution. In order to convert the mass of the solute into moles, you use its molar mass. The molar mass of sucrose is 342.2965 g/mol. Thus, the amount in moles is:

Moles sucrose = 3.68 g * (1 mol sucrose/342.2965 g) = 0.01075 moles

Next, we convert the volume of the solution from milliliters to liters. 1 liter is equivalent to 1,000 mL

Volume = 275 mL * (1 L/1000 mL) = 0.275 L

Hence, the molarity of the sucrose solution is
Concentration = 0.01075 moles/0.275 L
Concentration = 0.0391 M

The molarity of the solution is 0.03 M.

Molarity of a solution can be defined as the moles of solute present in a liter of the solution.

[tex]\rm Molarity\;=\;\dfrac{weight}{molecular\;weight}\;\times\;\dfrac{1000}{Volume\;(ml)}[/tex]

In the given question,

weight of sucrose = 3.68 g

Volume of solution = 275 ml

We know, the molecular weight of sucrose = 342 grams/mol

[tex]\rm Molarity\;=\;\dfrac{3.68}{342}\;\times\;\dfrac{1000}{275}[/tex]

Molarity = 0.03 M.

The molarity of the solution is 0.03 M.

For more information about molarity, refer to the link:

https://brainly.com/question/8732513?referrer=searchResults

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