Answer:
62.586 gram
Explanation:
moles of Al(OH)3 = mass / molar mass = 35.1 / (27+17x3) = 0.45 mol
moles of H2SO4 = mass / molar mass = 53.94 / (2+32+16x4) = 0.55 mol
H2SO4 is the limiting reagent (reacts completely)
⇒ moles of Al2(SO4)3 is worked out by moles of H2SO4
moles of Al2(SO4)3 = moles of H2SO4 / 3 = 0.183 mol
mass of Al2(SO4)3 = mole x molar mass = 0.183 x (27x2 + 96x3) = 62.586 gram
First, determine the amount of substance in moles by dividing the given amount in grams by the molar mass.
17.8 grams LiF x (1 mole / 25.94 grams LiF) = 0.686 moles LiF
Divide this amount of substance by the volume of the solution in liters.
molarity = 0.686 moles LiF / 0.915 L sol'n
The molarity is therefore, 0.75 M.