Percent by Weight Concentration of NaCl Solution
In a solution made by dissolving 7.60 g NaCl in enough water to give 87.8 g of solution, the percent by weight concentration is calculated as follows:
Percent by weight (% w/w) = (mass of solute / mass of solution) * 100%
% w/w = (7.00g / 87.39g) * 100%
% w/w = 8.7056%
Mass of NaCl Required for IV Solution
For a 0.92% w/v NaCl solution to be administered intravenously, the mass of NaCl required to prepare 345 mL of the solution is calculated as follows:
% w/v = (mass of solute / volume of solution) * 100%
mass of NaCl = (0.92% * 345 mL) / 100
mass of NaCl = 3.174 grams
Molarity of Different Solutions
a) For 2.37 moles of KNO3 dissolved in enough water to give 650 mL of solution, the molarity is calculated as follows:
Molarity (M) = moles of solute / volume of solution
M = 2.37 moles / 0.650 L
M = 3.6442 mol/L
b) For 250 g of NaOH dissolved in enough water to give 2.50 L of solution, the molarity is calculated as follows:
Molarity (M) = (mass of solute / molar mass of solute) / volume of solution
M = (250 g / 40.0 g/mol) / 2.50 L
M = 2.509 mol/L
Formality of Solutions
a) For a 28.4% NH3 by weight solution with a density of 0.808 g/mL, the formality is calculated as follows:
Formality (F) = moles of solute / volume of solution
F = 0.2008 mol/L
b) For a 36.0% HCl by weight solution with a density of 1.19 g/mL, the formality is calculated as follows:
Formality (F) = moles of solute / volume of solution
F = 11.7531 mol/L
Mole Fraction, Normality, and Molality
a) The mole fraction of NH3 in a 28.4% NH3 solution is calculated to be 0.2950.
b) The normality of a solution can be calculated using the formula N = M x f.
c) The molality of a solution is calculated as m = (moles of solute / kg solvent).
Equivalent Weight Calculations
a) The equivalent weight of Na3PO4 is calculated to be 54.6667 g/eq.
b) The equivalent weight of Al(OH)3 is calculated to be 16.3333 g/eq.
Normality Calculation
The normality of a solution resulting from the dissolution of 4.00 g of Al(NO3)3 in 250.0 mL is calculated as approximately 0.2254 N.
