1 mole
72.0 grams NaCl (1 mole NaCl/58.44 grams) = 1.2 moles of sodium chloride
1.946 moles NaCl (58.44 grams/1 mole NaCl) = 113.7 grams sodium chloride ======================
Divide it by molar mass. So the answer is 0.518 mol
molecular formula for sodium chloride = NaClIf the mole (n) for NaCl = 5.3 moles, then the mole of sodium (Na) = 5.3 moles as well. 1 to 1 ratio mass = moles X molar mass m = 5.3 x 22.9 = 121.37 grams of sodium in 5.3 moles of sodium chloride
3.0 moles NaCl [1 mole Na(+)/1 mole Cl(-)][6.022 X 10^23/ 1 mole Na(+)] = 1.8 X 10^24 sodium ions
Molarity = moles of solute/Liters of solution ( 300 ml = 0.300 Liters ) For our purposes, Moles of solute = Liters of solution * Molarity Moles NaCl = 0.300 Liters * 0.15 M = 0.05 moles NaCl =============
Based on the stoichiometry of NaCl, for every one mole of NaCl there is one mole of Na+ and one mole of Cl-. Therefore, there are 2.5 moles Na+ and 2.5 moles Cl-, totaling 5 moles of ions altogether.
Based on the stoichiometry of NaCl, for every one mole of NaCl there is one mole of Na+ and one mole of Cl-. Therefore, there are 1.5 moles Na+ and 1.5 moles Cl-, totaling 3 moles of ions altogether
Formal set up. 10 grams sodium chloride (1 mole NaCl/58.44 grams)(1 mole Na/1 mole NaCl)(6.022 X 1023/1 mole Na) = 1.0 X 1023 atoms of sodium --------------------------------------
1 mole NaCl = 58.443g NaCl 234g NaCl x 1mol NaCl/58.443g NaCl = 4.00 moles NaCl
79,9 g NaCl is equivalent to 1,367 moles.The formula mass of the compound sodium chloride, NaCl is 23.0 + 35.5 = 58.5Amount of NaCl = 79.9/58.5 = 1.37molThere are 1.37 moles of sodium chloride in a 79.9 gram pure sample of the compound.
For this you need the atomic mass of Na. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel.11.5 grams Na / (23.0 grams) = .500 moles Na