How many grams of nitric acid are present in 250.0 mL of 6.70 M acid solution?

1 Answer
Feb 10, 2016

Approx. 100 g.

Explanation:

"Concentration", mol*L^-1 = "Moles(n)"/"Volume of solution(L)".

Clearly, n, "No. of moles" = "Concentration "xx" Volume".

So, 250xx10^(-3)cancelLxx6.70*mol*cancelL^-1 = ?? mol HNO_3.

And now, multiply that molar quantity by the molar mass of nitric acid, 63.01*g*mol^-1.

?? mol xx 63.01*g*mol^-1 = ?? g?