The molecules of the solvent are combining, cutting the concentration of the solvent in half. This, therefore, doubles the molar mass of the solute. Example: 1. Suppose the freezing point of a solution of 2.00 g of an unknown molecular substance in 10.00 g of the solvent benzene is measured. If the solution freezes at a temperature of 6.33°C lower than pure benzene itself, calculate the molar mass of the unknown substance. - 6.33/5.12= 1.24m, solve for x- 1.24m= x/0.01 kg, x=0.0124 mol, 2.00 g/0.0124mol=161.29 g- your molar mass - now cut the amount of solvent, 0.01Kg(10.00g), in half 1.24m=x/0.005 kg, x=0.0062 mol, 2.00 g/0.0062mol = 322.581 g, your larger molar mass
The molar mass would increase with the forming of a dimer because the dimer decreases pressure. If you look at the ideal gas equation you will see that pressure is in the denominator, thus leading to an increase in molar mass.
n°moles solute = 5.3*0.125 = 0.6625 molar mass = 62.0/0.6625 = 93.58 g/mol
I'm trying to figure out how it affects molar mass as well. However, I do know that it decreases the pressure which is in the denominator of the equation to find the molar mass once the ideal gas equation is manipulated. So i think that with the formation of dimers the molar mass would increase?
This is the ratio between the molar concentration of this solute and 3,3.
150amu
The value is 2 molar.
This is the ratio concentration of a solute/molar mass of the solute.
number of moles=solute concentration/solute molar mass
Yoy need to know the molar mass of this solute and the mass of the solute:moles = total mass/molar mass
This is the molar concentration of the solute.
n°moles solute = 5.3*0.125 = 0.6625 molar mass = 62.0/0.6625 = 93.58 g/mol
The concentration of the solute is 0,5 molar.
I'm trying to figure out how it affects molar mass as well. However, I do know that it decreases the pressure which is in the denominator of the equation to find the molar mass once the ideal gas equation is manipulated. So i think that with the formation of dimers the molar mass would increase?
This is the ratio between the molar concentration of this solute and 3,3.
150amu
The molecular weight. So you'll have to calculate that for the solute first. The molar mass of the solute, which is measured in grams/mole.
The value is 2 molar.
This is the molar fraction.