A stock solution of HNO₃ is prepared and found to contain 13.7 mol L^-1 of . If 25.0 mL of the stock solution is diluted to a final volume of 0.500 L, what is the concentration of the diluted solution?

Question

Quizplus · Accepted Answer

First, let's use the given information to find the number of moles of HNO₃ in the 25.0 mL sample: moles of HNO₃ = (13.7 mol L^-1 / 1 L) x (0.0250 L) = 0.3425 mol L^-1 Next, we can use the dilution equation to find the concentration of the diluted solution: M1V1 = M2V2 where M1 is the initial concentration (in mol L^-1/L) and V1 is the initial volume (in L), M2 is the final concentration and V2 is the final volume. Plugging in the given values, we get: (13.7 mol L^-1 / 1 L) x (0.0250 L) = M2 x (0.500 L) Solving for M2, we get: M2 = 0.685 mol L^-1 / L Therefore, the concentration of the diluted solution is 0.685 mol L^-1 / L, which corresponds to answer choice E.

A Stock Solution of HNO3 Is Prepared and Found to Contain

A Stock Solution of HNO₃ Is Prepared and Found to Contain