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)
Answer Details
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