A glucose (C6H12O6) solution is analyzed to confirm its concentration. A 25.0 ml sample, with a mass of 26.208 g, is evaporated to dryness. If the solid glucose residue has a mass of 1.310 g, find a) the mass percent and b) the molal concentration of the glucose solution.
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Percent mass is equal to the mass of a part divided by the mass of the whole time 100. In this problem it would be the mass of glucose divided by the mass of the glucose solution times 100.
Mass of glucose = 1.1310 g
Mass glucose solution = 26.208 g
Percent mass of glucose in the solution = 1.310 / 26.208 * 100 = 4.998%, or roughly 5% glucose by mass.
Molar concentration (or molarity) is equal to the number of moles of glucose divided by liters of solution.
We are told that there are 1.310 g of glucose. We are also given the chemical formula of glucose (C6H12O6) and from there we can calculate the number of moles of glucose we have.
1.310g of glucose / 180.16 g/mol = .007271 moles of glucose
We are told that the volume of the solution is 25 ml, which we can easily convert to liters
25ml = .025 L
Molarity = 0.007271 mol Glucose / .025 L = 0.29 mol/L glucose
So the molar concentration of the glucose solution is 0.29 M glucose.
Mass of glucose = 1.1310 g
Mass glucose solution = 26.208 g
Percent mass of glucose in the solution = 1.310 / 26.208 * 100 = 4.998%, or roughly 5% glucose by mass.
Molar concentration (or molarity) is equal to the number of moles of glucose divided by liters of solution.
We are told that there are 1.310 g of glucose. We are also given the chemical formula of glucose (C6H12O6) and from there we can calculate the number of moles of glucose we have.
1.310g of glucose / 180.16 g/mol = .007271 moles of glucose
We are told that the volume of the solution is 25 ml, which we can easily convert to liters
25ml = .025 L
Molarity = 0.007271 mol Glucose / .025 L = 0.29 mol/L glucose
So the molar concentration of the glucose solution is 0.29 M glucose.