What is the molarity of an aqueous solution containing 27.3 g of glucose (C6H12O6) in 1.7 L of solution?

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To find the molarity of the aqueous solution, we need to calculate the moles of glucose (C6H12O6) and then divide it by the volume of the solution.

Step 1: Calculate the moles of glucose (C6H12O6)
To determine the number of moles of glucose, we need to use its molar mass. The molar mass of glucose is calculated by summing the atomic masses of its constituent atoms.

Glucose (C6H12O6) contains:
6 carbon (C) atoms with a molar mass of 12.01 g/mol: 6 x 12.01 = 72.06 g/mol
12 hydrogen (H) atoms with a molar mass of 1.008 g/mol: 12 x 1.008 = 12.096 g/mol
6 oxygen (O) atoms with a molar mass of 16.00 g/mol: 6 x 16.00 = 96.00 g/mol

The molar mass of glucose (C6H12O6) = 72.06 g/mol + 12.096 g/mol + 96.00 g/mol = 180.156 g/mol

Given that there are 27.3 g of glucose, we can calculate the moles by dividing the mass by the molar mass:

Moles of glucose (C6H12O6) = 27.3 g / 180.156 g/mol ≈ 0.1516 mol

Step 2: Calculate the molarity of the solution
Molarity is defined as the number of moles of solute per liter of solution. In this case, the solute is glucose (C6H12O6), and the volume of the solution is 1.7 L.

Molarity (M) = Moles of solute / Volume of solution

Molarity (M) = 0.1516 mol / 1.7 L ≈ 0.0892 M

Therefore, the molarity of the aqueous solution containing 27.3 g of glucose (C6H12O6) in 1.7 L of solution is approximately 0.0892 M.