A student prepared a solution of sodium carbonate by adding 2.285g of the solid to a 250mL volumetric flask and adding water to the mark. Some of this solution was transfered to a buret. What volume of solution should the student transfer into a flask to obtain 3.69 mmol Na+? Answer in mL.
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(1) Calculate moles of Na2CO3
(2) Calculate molarity of Na2CO3 solution.
(3) Calculate volume needed.
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(1) 2.285 g x 1 mol / 106 g = 0.021557 mol Na2CO3
(2) M = n/V = 0.021557 mol / 0.250 L
M = 0.08623 mol / L
(3)
(a) 3.69 mmol = 3.69 x 10^-3 mol
(b) 3.69 x 10^-3 mol Na+ x 1 Na2CO3 / 2 Na+ = 0.001845 mol Na2CO3
(c) 0.001845 mol Na2CO3 x 1 L / 0.08623 mol = 0.02140 L
0.02140 L = 21.4 mL
(2) Calculate molarity of Na2CO3 solution.
(3) Calculate volume needed.
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(1) 2.285 g x 1 mol / 106 g = 0.021557 mol Na2CO3
(2) M = n/V = 0.021557 mol / 0.250 L
M = 0.08623 mol / L
(3)
(a) 3.69 mmol = 3.69 x 10^-3 mol
(b) 3.69 x 10^-3 mol Na+ x 1 Na2CO3 / 2 Na+ = 0.001845 mol Na2CO3
(c) 0.001845 mol Na2CO3 x 1 L / 0.08623 mol = 0.02140 L
0.02140 L = 21.4 mL