2. a solution is made by adding 1.23 mol of kcl to 1000.0 g of water. assume that the density of water (and the resulting solution) is 1.00 g/ml. a. calculate the formula weight of kcl. b. calculate the mass of kcl in grams. c. calculate the percent by mass of kcl in this solution. d. calculate the molarity of the solution.

a. 74.55 g/mol.

b. 91.70 g.

c. 8.40%.

d. 1.23 mol/L.

Explanation:

a. Calculate the formula weight of KCl.

∵ Formula weight of KCl = atomic weight of K + atomic weight of Cl

atomic weight of K = 39.098 g/mol, atomic weight of Cl = 35.45 g/mol.

∴ Formula weight of KCl = atomic weight of K + atomic weight of Cl = 39.098 g/mol + 35.45 g/mol = 74.548 g/mol ≅ 74.55 g/mol.

b. Calculate the mass of KCl in grams.

no. of moles (n) = mass/molar mass.

∴ mass of KCl = n*molar mass = (1.23 mol)*(74.55 g/mol) = 91.69 g ≅ 91.70 g.

c. Calculate the percent by mass of KCl in this solution.

The mass % of KCl = (mass of KCl/mass of the solution) * 100.

mass of KCl = 91.70 g,

mass of the solution = 1000.0 g of water + 91.70 g of KCl = 1091.70 g.

∴ The mass % of KCl = (91.70 g/1091.70 g)*100 = 8.399% ≅ 8.40%.

d. Calculate the molarity of the solution.

Molarity is the no. of moles of solute per 1.0 L of the solution.

M = (no. of moles of KCl)/(Volume of the solution (L))

no. of moles of KCl = 1.23 mol,

Volume of the solution = mass of water / density of water = (1000.0 g)/(1.00 g/mL) = 1000.0 mL = 1.0 L.

M = (1.23 mol)/(1.0 L) = 1.23 mol/L.

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.