Because a solution with impurities has a lower volatility. The mole fraction of the solvent is also decreased.
Molality = moles solute per kg solventmoles solute = 6 moles CaCl2kg solvent = 3 kg watermolality = 6 mole/3 kg = 2 m
Mole fraction and mole percent are essentially describing the same quantity, only mole percent uses the conversion of a number to a percentage. That's all the difference there is. Hence, a mole fraction of 0.134 for a component in a mixture is equivalent to 13.4 % in mole percent of that component. Mole percent spans the range from 0 to 100% while mole fraction spans the real values from 0 to 1.
Mole fraction to volume fraction could be converted using the molar mass and the density of the particles involved in the liquid.
30.1 percent by mass = 30.1 g HCOOH + 69.9 g H2Omoles HCOOH = 30.1 g x 1 mol/46 g = 0.654 molesmoles H2O = 69.9 g x 1 mol/18 g = 3.88 molesTotal moles = 4.53mole fraction HCOOH = 0.654/4.53 = 0.144mole fraction H2O = 3.88/4.53 = 0.856molality = moles solute/kg solvent = 0.654/0.0699 = 9.36 m
No. The mole of solution is equal to mole of solute plus mole of solvent (solVENT !, not solution)
number of moles of solute/solvent divided by number of moles of the solution
The molar mass of benzene (C6H6) is 78.11 grams per mole. The molar mass of toluene is 92.14 grams per mole. Mole fraction of a solute equals moles of solute divided by the moles of solute plus the moles of solvent. .5 of each is the mole fraction.
(mole of solute/mole of solvent)*100
molal = m = moles solute/kg solvent. Assuming water as the solvent 1 kg = 1000 g/18g/mole = 55.56 moles H2O. Moles solute = 3.71 moles. Total moles present = 55.56 + 3.71 = 59.27 moles.Mole fraction solute = 3.71/59.27 = 0.0626
Mass percent = grams of solute/total grams of solution Mole fraction = mols component/total mols mix. Molarity = mols solute/L solution Molality = mols solute/kg solvent Hope this helps :)
The volume fraction of a substance is equal to the mole fraction for ideal gas mixture
The law which states that vapor pressure of solvent decreases in proportion to the concentration of the solute is Raoult's law .This law states that the partial vapor pressure of a component in a mixture is equal to the vapor pressure of the pure component at that temperature multiplied by its mole fraction in the mixture.
0.0613 mole fraction = moles solute/(moles solute + moles water) moles water = 1000 g / 18.02 = 55.49 3.62/(55.49 + 3.62)=0.0613
This is the ratio between the molar concentration of this solute and 3,3.
Pa= mole fraction x pressure of the pure solvent