Determine the mass (in g) of nitric acid (HNO3) in 60.00 cm3 of a 3.00 mol dm–3 solution. Answer: Mass of HNO3 = g (to three significant figures).

Sure, here is the step by step solution:

1. First, we need to convert the volume from cm3 to dm3 because the concentration is given in mol/dm3. We know that 1 dm3 = 1000 cm3. So, 60.00 cm3 = 60.00/1000 = 0.0600 dm3.

2. Next, we use the formula for concentration which is moles/volume. We rearrange this to find moles = concentration x volume. Substituting the given values we get moles of HNO3 = 3.00 mol/dm3 x 0.0600 dm3 = 0.180 mol.

3. Now, we need to convert moles to grams. We know that 1 mole of HNO3 = 63.01 g (1H) + 14.01 g (1N) + 48.00 g (3O) = 63.01 + 14.01 + 48.00 = 125.02 g.

4. So, the mass of HNO3 = moles x molar mass = 0.180 mol x 125.02 g/mol = 22.504 g.

5. To three significant figures, the mass of HNO3 = 22.5 g.