a 35.0-mL sample of 0.513 M glucose (C6H12O6) solution is mixed with 140.0 mL of

Question

a 35.0-mL sample of 0.513 M glucose (C6H12O6) solution is mixed with 140.0 mL of 2.33 M glucose solution. what is the concentration of the final solution? Assume the volumes are additive. a 35.0-mL sample of 0.513 M glucose (C6H12O6) solution is mixed with 140.0 mL of 2.33 M glucose solution. what is the concentration of the final solution? Assume the volumes are additive. a 35.0-mL sample of 0.513 M glucose (C6H12O6) solution is mixed with 140.0 mL of 2.33 M glucose solution. what is the concentration of the final solution? Assume the volumes are additive.

Explanation / Answer

Number of moles = molarity * volume of solution in L

Number of moles = 0.513 * 0.035 = 0.0180 mole

Number of moles =  molarity * volume of solution in L

Number of moles = 2.33 * 0.140 = 0.326 mole

total number of moles in the solution = 0.0180 + 0.326 = 0.344 mole

volumes are additive = 35.0 + 140.0 = 175 mL = 0.175 L

Therefore, the concentration of the final solution = total number of moles/ volume

the concentration of the final solution = 0.344 mol / 0.175 L = 1.97 M

Therefore, the concentration of the final solution = 1.97 M

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