G What is the molarity of an aqueous solution that is 2.50 % glucose (C6H12O6) by mass? Assume that the density of the solution is 1.04 g/cm3

0.144 M

Explanation:

First, we will calculate the mass/volume percent (% m/v) using the following expression.

% m/v = % m/m × density

% m/v = 2.50 % × 1.04 g/cm³

% m/v = 2.60 %

This means that there are 2.60 grams of solute per 100 cm³ (100 mL) of solution. The molarity of glucose is:

M = mass of glucose / molar mass of glucose × liters of solution

M = 2.60 g / 180.16 g/mol × 0.100 L

M = 0.144 M

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