An aqueous solvent of NaCl has a melting point of –165C
How many moles of NaCl are present in a 189 g solvent?
I don't believe your numbers. -165C as the freezing point is a pretty great amount of salt.
freezingpoint=molarity*-.52
molarity=165/.52 about 300 moles/kg solvent
molesNaCl=about 300/.189 about 1600 moles salt or about more than a barrel full of salt in a beaker of water? It isn't going to happen, at least here in Texas.
Actually it is even worse than that since Kf is 1.86 (and not 0.52)
True.
To determine the number of moles of NaCl present in a solvent, we need to use the formula:
moles = mass / molar mass
First, let's find the molar mass of NaCl. The molar mass is the sum of the atomic masses of sodium (Na) and chlorine (Cl). The atomic mass of sodium is approximately 22.99 g/mol, while the atomic mass of chlorine is approximately 35.45 g/mol. Adding them together:
molar mass of NaCl = (22.99 g/mol) + (35.45 g/mol) = 58.44 g/mol
Now, we can calculate the number of moles of NaCl using the given mass of the solvent:
moles = 189 g / 58.44 g/mol
moles ≈ 3.24 mol
Therefore, there are approximately 3.24 moles of NaCl present in a 189 g solvent.