delta T = Kf*molality
Plug in 5.00 C for delta T, 1.86 for Kf and calculate molality.
Then molality x kg solvent x molar mass = grams.
You know molality, kg solvent, and molar mass. Calculate grams.
molar mass? where did u got the molar mass from ?
You want grams, don't you?
The formula I have came from this.
molality = mols/kg solvent. Solve for mols.
mols = molality x kg solvent.
mols also = grams/molar mass
so we set mols = mols
so molality x kg solvent = g/molar mass
multiply both sides by molar mass an we have
molality x kg solvent x molar mass = grams.
You didn't copy your question but it was something with an OH on/in it. That's what you want the molar mass of. I think I remember the question as being determine the mass of the material to make the freezing point decrease to -5.00 C.
That's ethylene glycol
No. I went back and reread the problem. It asks for mols, not grams.
so molality = mols/kg solvent and this can be rearranged to
molality x kg solvent = mols
You know m and you know kg solvent. Solve for mols.
Apologies for the confusion. In order to calculate the grams, we need to know the molar mass of the solute. The molar mass represents the mass of one mole of a substance and is usually expressed in units of grams per mole (g/mol). It is determined by adding up the atomic masses of all the atoms present in the chemical formula of the substance.
To find the molar mass, you can use the periodic table which lists the atomic masses of all the elements. Multiply the atomic mass of each element by the number of atoms of that element in the chemical formula, and then add up the values for all the elements.
For example, if the solute is glucose (C6H12O6), you would find the atomic masses for carbon (C), hydrogen (H), and oxygen (O) on the periodic table. Then multiply the atomic mass of carbon by 6, hydrogen by 12, and oxygen by 6, and finally add up the results to find the molar mass of glucose.
Once you have the molar mass, you can proceed with the calculations as mentioned earlier to find the grams of the solute.