Problem 72P

Calculate the mass of glucose (C6H12O6) in a 105-mL sample of a 1.02 M glucose solution.

Solution: here, we are going to calculate the mass of glucose(C6H12O6) in the sample solution.

Step1:

Molarity(M) is defined as number of moles of solute dissolved in one litre (or one cubic decimetre) of solution.

Moles of solute

Molarity = ------------------------------- ------(1)

Volume of solution in litre

Step2:

In the above problem,

Volume of the solution = 105 mL = 105 / 1000 L = 0.105 L [1000 mL = 1 L]

Molarity of the solution = 1.02 M

Substituting the values in equation (1), we get,

1.02 M = moles of C6H12O6 / 0.105 L

Therefore, moles of C6H12O6 = 1.02 M x 0.105 L

= 0.1071 mol

Step3:

1 mol C6H12O6 = molar mass of C6H12O6

0.1071 mol C6H12O6 = 0.1071 x molar mass of C6H12O6

= 0.1071 x 180.16 g/mol

= 19.295 g.

Thus, the mass of C6H12O6 is 19.295 g.

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