Question

A 65.0 mL sample of 0.513 M glucose (C6H12O6) solution is mixed with 160.0 mL of 2.33 M glucose solution. What is the concentration of the final solution? Assume the volumes are additive.

Answer 1

NO. Of moles of Glucose in solumon 1 = Moloriry x volume (L) = 0.513 mol x 6500 ml x 0.001L Iml = 3.33 x10² moe • NO of molus of Gwcost IN SOLUTION 2 = 2.33 x 160.0 mL X 0.001L ImL - 0.3728 mol TOTAL No. of moles of Glucose = 3.33x10² mol + 0.3728 more = 0.406, moe e volume of SoutON = 6500 mL + 16010 ml = 22510 ml CONCENTRATION OF FINAL sowion = 0.4001 moe -1).80 M 22500 mL x 0.001L Ime