This is a sodium chloride solution with the concentration of 0,5844 g NaCl to
1 L water.
The concentration of sodium chloride would be 1.5 moles per liter (M). If 200 ml is used, that would be 0.3 moles of sodium chloride.
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)
To make a 1.0 M solution of sodium cation using sodium chloride, you need to consider the molar mass of sodium chloride (58.44 g/mol). Since 1 mole of sodium chloride provides 1 mole of sodium cation, you would need 58.44 g of sodium chloride to make a 1.0 M solution in 100 ml.
The answer is 0,1648 g NaCl.
0,6 M of sodium chloride is 35,064 g.
In chemistry, the concentration of a substance in solution is determined by molarity, which is symbolized by "M". This indicates the number of moles of a substance dissolved in one liter of a solvent (usually water). For example: - 1 mole of sodium chloride = 58 grams - If 116 grams of sodium chloride are dissolved in 1 liter of water, then that solution is a 2-molar (2 M) solution of sodium chloride. - If 232 grams of sodium chloride are dissolved in 1 liter of water, then that solution is a 4-molar (4 M) solution of sodium chloride.
To find the number of moles of sodium chloride in a 250 mL solution with a concentration of 1.20 m, you can use the formula: moles = concentration (mol/L) × volume (L). First, convert 250 mL to liters, which is 0.250 L. Then, multiply: 1.20 mol/L × 0.250 L = 0.30 moles of sodium chloride.
The needed mass of sodium chloride is 29,22 g.
To find the volume of a 0.62 M solution needed to obtain 0.02 moles of sodium chloride, you can use the formula: [ \text{Volume} = \frac{\text{Number of moles}}{\text{Concentration}} = \frac{0.02 \text{ moles}}{0.62 \text{ M}} \approx 0.03226 \text{ L} \text{ or } 32.26 \text{ mL}. ] Therefore, you would need approximately 32.26 mL of the 0.62 M sodium chloride solution.
The first step is to convert the mass of sodium chloride to moles using its molar mass. Then, divide the moles of sodium chloride by the volume of the solution in liters to calculate the molarity.
Silver nitrate is not soluble in a sodium chloride solution.
The osmotic pressure is 24,44 at.