How many grams of H2SO4 are necessary for the preparation of 0.175 dm3 of M 6.00 M H2SO4? (S = 32.06, O = 16.00 H = 1.00)

To calculate the grams of H2SO4 needed for the preparation of 0.175 dm³ of 6.00 M H2SO4, we need to use the formula: Molarity (M) = moles (n) / volume (V) Rearranging the formula, we get: moles (n) = Molarity (M) x volume (V) We are given the volume (V) as 0.175 dm³ and the Molarity (M) as 6.00 M. So, moles of H2SO4 = 6.00 M x 0.175 dm³ = 1.05 moles of H2SO4 The molar mass of H2SO4 can be calculated as: 2(1.00 g/mol) + 32.06 g/mol + 4(16.00 g/mol) = 98.08 g/mol Therefore, 1.05 moles of H2SO4 weighs: 1.05 moles x 98.08 g/mol = 102.98 g So, approximately 103.0 grams of H2SO4 are necessary for the preparation of 0.175 dm³ of 6.00 M H2SO4. Therefore, the answer is: 103.0 g