Glucose (molar mass=180.16 g/mol) is a simple, soluble sugar. Glucose solutions are used to treat patients with low blood sugar. Suppose you prepare a glucose solution using the described procedure. Step 1: Dissolve 263.1 g of glucose in enough water to make 500.0 mL of solution. Step 2: Transfer 19.2 mL of the solution to a new flask and add enough water to make 250.0 mL of dilute solution. What is the molar concentration of the glucose solution at the end of the procedure?

Molar mass (glucose): 6×12.0011 g+12×1.00794 g+6×15.9994 g=180.158 g,1.5×10−1mol ×180.158 g/mol =27 g. 6 × 12.0011 g + 12 × 1.00794 g + 6 × 15.9994 g = 180.158 g , 1.5 × 10 − 1 mol × 180.158 g/ mol = 27 g.wer:

Explanation:

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dsdrajlin dsdrajlin · Answer 2 · 2021-09-03T20:38:05+02:00

The molar concentration of the glucose solution at the end of the procedure is 0.224 M.

In the first step, we dissolve 263.1 g of glucose (molar mass=180.16 g/mol) in enough water to make 500.0 mL of solution. We can calculate the molarity of the initial solution using the following expression.

[tex]M = \frac{mass\ glucose}{molar\ mass\ glucose \times liters\ solution } = \frac{263.1 g}{180.16 g/mol \times 0.5000 L } = 2.921 M[/tex]

In the second step, we take a volume (V₁) of 19.2 mL of the 2.921 M (C₁) solution and add enough water to make 250.0 mL (V₂) of dilute solution. We can find the concentration of the dilute solution (C₂) using the dilution rule.

[tex]C_1 \times V_1 = C_2 \times V_2\\C_2 = \frac{C_1 \times V_1}{V_2} = \frac{2.921 M \times 19.2 mL}{250.0 mL} = 0.224 M[/tex]

The molar concentration of the glucose solution at the end of the procedure is 0.224 M.

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