Solution A is prepared by dissolving 45 g of glucose ( C ₆ H ₁₂ O ₆ ) in water to make 500 mL of solution. Solution B is a 0.2 M aqueous solution of glucose. A volume of 200 mL of Solution A is mixed with 300 mL of Solution B, and enough wa

Question

Solution A is prepared by dissolving 45 g of glucose ( C ₆ H ₁₂ O ₆ ) in water to make 500 mL of solution. Solution B is a 0.2 M aqueous solution of glucose. A volume of 200 mL of Solution A is mixed with 300 mL of Solution B, and enough water is added to the mixture to bring the final volume to 800 mL . The molarity of the final solution is x 10⁻² M . The value of x is. Given: Atomic masses of H , C , and O are 1 ,

Quantrex Academy · Accepted Answer

Correct answer: 20

First, calculate the molarity of Solution A. Molar mass of glucose ( C ₆ H ₁₂ O ₆ ) = 6(12) + 12(1) + 6(16) = 180 g mol ⁻¹ Moles of glucose in 500 mL of Solution A = 45 g 180 g mol ⁻¹ = 0.25 mol Molarity of Solution A ( M_A ) = 0.25 mol 0.5 L = 0.5 M Next, calculate the number of millimoles of glucose contributed by the volumes of A and B mixed: Millimoles from Solution A = M_A V_A = 0.5 M 200 mL = 100 mmol Millimoles from Solution B = M_B V_B = 0.2 M 300 mL = 60 mmol Total millimoles of glucose in the mixture = 10

Correct answer

Step-by-step solution

More from Solutions