Calculating mass percentage, mole fraction, molality, as well as molarity?
(a) Calculate the mass percentage of ascorbic acid in this solution. 82.1 grams / 210 g = 39.1% ----------------------- (b) Calculate the mole fraction of ascorbic acid in this solution. 82.1 g Ascorbic @ 176.13 g/mol = 0.466 moles 210 g water @ 18.0 g/mol = 11.666 moles water 0.466 & 11.666 = 12.133 moles total 0.466 moles ascorbic / 12.1 total = 0.0384 mole fraction ascorbic ------------------ (c) Calculate the molality of ascorbic acid in this solution. 0.466 moles / 210 grams times 1000 = 2.22 molal --------------------------- (d) Calculate the molarity of ascorbic acid in this solution solution weighs 82.1 plus 210 = 292.1 grams 292.1 grams @ 1.22 g/ml = 239.4 millilitres 0.466 moles / 0.2394 litres = 1.95 Molar
Calculating molality if given mole fraction?
Assume you have 1 total mole in the solution acetone moles = 0.221 water moles = 0.779 acetone g = 0.221(58 g/mole) = 12.8g water g = 0.779(18 g/mole) 14.0 g total weight = 26.8 g = 0.026kg molality = 0.221 moles/0.026kg = 8.5m
How do you use molality to calculate mole fraction?
Well, look at the defining relationships….[math]\text{Molality}=\dfrac{\text{Moles of solute}}{\text{Kilograms of solvent}}[/math]And ……[math]χ_{\text{the mole fraction}}=\dfrac{\text{Moles of solute}}{\text{Total moles in solution, of solutes and solvent}}[/math]And thus let us address the molality and mole fraction of a solution prepared from a [math]58.44•g[/math] mass of sodium chloride dissolved in a one litre volume (i.e. a one kilogram mass) of water…Now clearly, [math]\text{molality}=\dfrac{1•mol}{1•kg}=1•mol•kg^{-1}…[/math]But [math]χ_{NaCl}=\dfrac{1•mol}{1•mol+\frac{1000•g}{18.01•g•mol^{-1}}}=0.0177[/math]…It does not take too much mathematical nous to realize that [math]χ_{\text{water}}=1–0.0177=0.9823[/math]…because the moles fractions MUST sum to UNITY in a binary solution…Are you happy with this?
How do you relate mole fraction and molality?
The mole fraction is the ratio of moles of solute to moles of solvent. Molality is the number of moles of solute per mass of solvent.
Calculate the molality and mole fraction of 2.5g of ethanoic acid (CH3COOH) in 75g of benzene?
1.Mole fraction is defined as= Number of moles of solute/ Number of moles of solution.Molecular mass of ethanoic acid=60Molecular mass of benzene=78Number of moles of ethanoic acid= 2.5/60=0.041Number of moles of benzene= 75/78=0.96153Mole fraction= 0.041/(0.041+0.96153)=0.0402. Molality = Number of moles of solute/weight of the solvent (in kg)= 0.041/0.075=0.5466
Can you calculate the molarity, molality, and mole fraction of 28% H2SO4 solution (d=1.038g/cm3)?
grams of solute:in 1 Lm = 0.28 x 1.038 x 1000 = 290.64 gmole = 290.64 /98 g/mol = 2.97Molarity = 2.97 MMolallity?mass of solution = DV = 1.038 x 1000 = 1038 gmass of solvent: 1038 -290.64 = 747.36 = 747.4molaliti = 2.97 / 0.7474 =3.97mole fractionSolute: 2.97 molesolvent: 747.4 /18 = 41.52mole fraction2.97 / (2.97 + 41.52) = 2.97 /44.49 = 0.066842.52 / 44.49 = 0.9332
Calculate molality, molarity, and mole fraction of NH3 in 8.00 mass % aqueous solution (density =0.9561g/mL)?
8.00 x 956.1 g/L / 100 =76.5 g/L M = 76.5 g/L / 17.03 g/mol =4.49 mass water = 956.1 - 76.5 = 879.6 g = 0.8796 Kg => 879.6 g / 18.02 g/mol =48.8 moles m = 4.49 / 0.8796 = 5.10 total moles = 4.49 + 48.8 =53.29 mole fraction NH3 = 4.49 / 53.29 =0.0843
Finding the mole fraction from molality?
You are practically there. Mole fraction is a ratio of the moles of one substance to the total moles in the mixture. Molality is the ratio of the moles of the solute to 1 kilogram of the solvent. You can compute the moles of solvent. In a 1.44 molal solution of NaNO3 you have 1.44 moles of NaNO3, and ... 1000 grams H2O x 1 mol H2O / 18 g H2O = 55.6 mol water. So the mole fraction of NaNO3 is ... 1.44 / (1.44 + 55.6) = 0.0252