If 100.mL of 0.400 mol L^-1 Na₂SO₄(aq)is added to 200 mL of 0.600 mol L^-1 NaCl(aq),what is the concentration of Na⁺ ions in the final solution? Assume that the volumes are additive. A)0.534 mol L^-1 B)0.667 mol L^-1 C)1.00 mol L^-1 D)1.40 mol L^-1

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First, we need to calculate the moles of Na⁺ in the 100 mL of 0.400 M Na^-1 solution: moles of Na⁺ = volume (L) x concentration (mol/L) moles of Na⁺ = 0.1 L x 0.400 mol/L = 0.040 mol Next, we need to calculate the moles of NaCl in the 200 mL of 0.600 M NaCl solution: moles of NaCl = volume (L) x concentration (mol/L) moles of NaCl = 0.2 L x 0.600 mol/L = 0.120 mol Since both solutions are mixed and the volumes are additive, we now have a total volume of 0.1 L + 0.2 L = 0.3 L of solution. To find the overall concentration of Na⁺ ions in the solution, we need to add the moles of Na⁺ and NaCl together and divide by the total volume: total moles of Na⁺ and NaCl = 0.040 mol + 0.120 mol = 0.160 mol overall concentration of Na⁺ ions = total moles / total volume overall concentration of Na⁺ ions = 0.160 mol / 0.3 L = 0.533 mol/L, which is closest to 0.667 mol L^-1, so the answer is B.

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