This question can be answered with the equation n = m ÷ M (where n is the number of mol's, m is te mass in grams and M is the Molar Mass of Sodium Chloride). Now, sodium chloride's molecular formula is written as, NaCl Now if we think of this formula as representing a ratio of elements within the compound, we can see that, NaCl : Cl = 1 mole of sodium chloride molecules : 1 mole of Chloride atoms So, as we know the mol ratio for this molecule, we can establish the mass of sodium in 11.7g of sodium chloride by first calculating the number of moles of Sodium chloride in 11.7g of the substance, so then moles = mass of sodium chloride ÷ Molar mass of sodium chloride = 11.7 ÷ 58.44277 = 0.20019584971759552122529442050745 moles (note that in stochiometry i personally prefer not to round numbers until the end of the question Now by applying this number to the mol ratio, we can calculate the number of moles of sodium in sodium chloride 1 mole of sodium chloride molecules : 1 mole of Chloride atoms =0.20019584971759552122529442050745 : 0.20019584971759552122529442050745 Now that we have established that the number of moles of of sodium in 11.7g of sodium chloride is 0.20019584971759552122529442050745 mol, the mass of sodium in 11.7g of sodium chloride can be calculated with the formula, m = n M = mass of sodium in 11.7g of sodium chloride = 0.20019584971759552122529442050745 X 23 = 4.6045045435046969881817716716713g (all decimal places)
This value is 58,439 769 28.
26 g sodium chloride contain 10,313 g sodium.
The mass is 58,44 g NaCl.
The molecular mass of sodium chloride is 58,44 g.
The molecular weight of sodium chloride is 58,44 (rounded).
100 g of the solution contains 11 g of sodium chloride
The answer is 2080,46 g.
50 g sodium chloride (NaCl) contain 30,167 g chlorine.
The molecular mass of sodium chloride is 58,44 g.
The ratio mass of chlorine/mass of sodium is 1,5.
The molar mass of sodium chloride is the sum between the atomic weight of sodium and chlorine: 22,98976928 + 35,45 = 58,439 769 28
14.35 g + 8.5 g - 5.85 g or 17 g.
The answer is 1,105 g.