moles NaOH = c · V = 0.1973 mmol/mL · 29.43 mL = 5.806539 mmol
moles H2SO4 = 5.806539 mmol NaOH · 1 mmol H2SO4 / 2 mmol NaOH = 2.9032695 mmol
Hence
[H2SO4]= n/V = 2.9032695 mmol / 32.42 mL = 0.08955 M
The answer to this question is [H2SO4] = 0.08955 M
Be-beryllium have 2 electrons and it is in the 2 nd period
Mol sulfuric acid = 19 g * (1 mol) / (98.1 g) = 0.19367 mol
mol H2O = 0.19367 mol H2SO4 * (2 H2O) / (1 H2SO4)
= 0.387359 mol H2O
grams H2O = 0.387359 mol H2O * (18 g)/(1 mol)
= 6.97 g
The answer is 7.0 grams of water
