Final answer:
To determine the molarity of a 17.5% by mass aqueous nitric acid solution, moles of nitric acid are calculated assuming 100 g of solution and the volume approximated as 100 mL. The molarity is then found to be 2.778 M, with the closest given option being 3.0 M.
Step-by-step explanation:
To calculate the molarity of a 17.5% by mass aqueous solution of nitric acid (HNO3), we assume we have 100 g of the solution. This means we have 17.5 g of HNO3 and 82.5 g of water. The molar mass of HNO3 is approximately 63.01 g/mol.
First, calculate the number of moles of HNO3:
(17.5 g) / (63.01 g/mol) = 0.2778 moles HNO3
Next, find the volume of the solution. Since density is not provided, we approximate the density of the solution to be close to that of water, which is 1 g/mL due to its high percentage of water. So 100 g of solution is approximately 100 mL or 0.1 L.
Now, calculate the molarity:
(0.2778 moles) / (0.1 L) = 2.778 M
However, we don't have an option that matches 2.778 M exactly. The closest option is:
3.0 M