In the first problem, you are given a 7.58 gram sample of benzoic acid (C6H5CO2H) dissolved in 100 mL of benzene. The first step is to convert the volume of benzene to grams by multiplying it by the density of benzene, which is 0.879 g/mL. So, 100 mL * (0.879 g benzene/1 mL) equals 87.9 g of benzene.
Next, you need to convert the mass of benzene to kilograms by dividing it by 1000 (since there are 1000 grams in a kilogram). So, 87.9 g * (1 kg/1000 g) equals 0.0879 kg of benzene.
The molecular weight of benzoic acid is 122.12 g/mol. To find the number of moles of benzoic acid, divide the mass (7.58 grams) by the molecular weight: 7.58 g benzoic acid * (1 mol benzoic acid/122.12 g) equals 0.0621 mol benzoic acid.
Finally, to find the molality of the benzoic acid, divide the number of moles of benzoic acid by the mass of the solvent in kilograms: 0.0621 mol / 0.0879 kg = 0.706 mol/kg of benzene.
In the second problem, you are given that 15.48 mL of a Ba(OH)2 solution is required to react with 25 mL of a 0.303 M HCl solution. The balanced reaction is 2HCl(aq)+ Ba(OH)2(aq) -> BaCl2 + 2H2O(L).
To find the molarity of the Ba(OH)2 solution, you need to calculate the number of moles of HCl used in the reaction. Multiply the concentration (0.303 M) by the volume (25 mL converted to liters by dividing by 1000): 0.303 M * 0.025 L = 0.00758 mol HCl.
Since the coefficients of HCl and Ba(OH)2 in the balanced equation are not the same, you cannot use the formula m1v1 = m2v2 directly. Instead, you need to use stoichiometry to convert the moles of HCl to moles of Ba(OH)2. According to the balanced equation, for every 2 moles of HCl, 1 mole of Ba(OH)2 is needed.
So, 0.00758 mol HCl * (1 mol Ba(OH)2/2 mol HCl) = 0.00379 mol Ba(OH)2.
Finally, to find the molarity of the Ba(OH)2 solution, divide the number of moles (0.00379 mol) by the volume (15.48 mL converted to liters by dividing by 1000): 0.00379 mol / 0.01548 L = 0.245 M.
Therefore, the molarity of the Ba(OH)2 solution is 0.245 M.