Question

A 350. mL solution of 0.150 M NaNO₃(aq) is mixed with a solution of 230. mL of 0.240 M NaCl(aq). How many moles of Na⁺(aq) are present in the final solution?

            A)      0.0525 moles D)      0.0539 moles

            B)      0.108 moles E)      0.195 moles

            C)      0.186 moles   

Answer 1

Answer

B) 0.108moles

Explanation

Molarity = moles of solute per liter of solution

Number of moles of NaNO3 in 350ml solution = (0.150mol/1000ml) × 350ml = 0.0525mol

Number of moles of Na⁺ in 350ml solution = 0.0525mol

Number of moles of NaCl in 230ml solution = (0.240mol/1000ml) × 230ml = 0.0552mol

Number of moles of Na⁺ in 230ml = 0.0552mol

Total moles of Na⁺ = 0.0525mol + 0.0552mol = 0.108mol