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An aqueous solution of glucose (MM = 180.2 g/mol) has a molality of 2.27 m and a density of 1.20 g/mL. What is the molarity of glucose in the solution?

**Any help would be greatly appreciated!**

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mickymike92

Answer:

Molarity of the glucose solution = 2.72 mol/L

Explanation:

Molality of a solution is defined as the number of moles of solute per kilogram of solvent.

Therefore , a 2.27 molal solution of glucose contains 2.27 moles of glucose in 1 Kg of solvent

Density of solution = 1.20 g/mL

The volume occupied by 1 Kg or 1000 g solution = mass/density

Volume of 1000 g solution = 1000 g/1.20 g/ml = 833.3 mL

Number of moles of glucose present in the solution = 2.27 moles

Molarity = number of moles / volume(L)

volume of solution in litres = 883.3/1000 = 0.8333 L

Molarity = 2.27 moles /0.8333 l = 2.72 mol/L

Therefore, molarity of the glucose solution = 2.72 mol/L

The molarity of the glucose in the solution has been 2.72 mol/L.

Molality can be defined as the moles of the solute per kg of solvent.

The molality of glucose solution = 2.27 m

2.27 moles of glucose in 1000 grams of water.

Density has been the mass per unit volume. The density can be expressed as:

Density = [tex]\rm \dfrac{Mass}{Volume}[/tex]

The density of the given glucose solution = 1.20 g/ml

The volume of 1000 grams of water has been:

Volume = [tex]\rm \dfrac{Mass}{Density}[/tex]

Volume of 1000 grams water = [tex]\rm \dfrac{1000\;g}{1.20\;g/ml}[/tex]

Volume of 1000 g water = 833.3 ml

The molarity can be defined as the mass of solute per liter of the solution.

Molarity = [tex]\rm moles\;\times\;\dfrac{1000}{Volume\;(ml)}[/tex] ......(i)

The moles of glucose in the 833.3 ml solution have been 2.27 mol.

Substituting the values in equation (i):

Molarity = 2.27 mol [tex]\rm \times\;\dfrac{1000}{833.3\;ml}[/tex]

Molarity = 2.72 mol/L.

The molarity of the glucose in the solution has been 2.72 mol/L.

For more information about the molarity of the solution, refer to the link:

https://brainly.com/question/14707974