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The molar mass of chlorine is 35.45 g/mol. To calculate the molar mass of NaCl, multiply the subscript of each element times its molar mass, then add them together. If there is no subscript, it is understood to be 1. (1 ×22.990g/mol) + (1 ×35.45g/mol) = 58.44 g/mol. "58.44277 g/mol" Sodium chloride, also known as table salt, is an ionic ...
"0.1 moles NaCl" The thing to remember about any solution is that the particles of solute and the particles of solvent are evenly mixed. This essentially means that if you start with a solution of known molarity and take out a sample of this solution, the molarity of the sample will be the same as the molarity of the initial solution. In your case, you dissolve "58.44 g" of sodium chloride in ...
Enhanced with AI, our expert help has broken down your problem into an easy-to-learn solution you can count on. See Answer. Question: The molar mass of NaCl is 58.44 g/mol. Calculate the molarity of 21% (w/v) aqueous solution of NaC. The molar mass of NaCl is 5 8. 4 4 g / mol. Calculate the molarity of 2 1 % (w / v) aqueous ...
Chemistry questions and answers. A student analyzes a mixture of solid sodium chloride (NaCl molar mass 58.44g/mol) and sodium acetate (NaC2H2O2, molar mass 82.03g/mol) by titration with hydrochloric acid (HCI molar mass 36.46g/mol). Sodium chloride does not react with HCI. Sodium acetate reacts with HCl according to the following equation ...
You need to find the molar mass of Na and Cl. If you look straight into the periodic table, you can identify it immediately. Na = 23.0 g mol^-1 Cl = 35.5 g mol^-1 Step 3: Add the component parts up and it will give you the molar mass of the compound (NaCl). 23.0 + 35.5 = 58.5 Therefore, the molar mass of Sodium Chloride (NaCl) is 58.5 g mol^-1.
What minimum mass of NaCl (molar mass = 58.44 g/mol) must be added to 1.25 L of 0.44 M AgNO3(aq) to cause precipitation of AgCl (Ksp of AgCl is 1.77x10-10)? Answer: 2.9x 10 g 8. What is the equilibrium concentration of silver ions when 100. mL of 0.200 M Na3PO4(aq) is combined with 25.0 mL of 0.300 M AgNO3(aq)?
Step 1 of 3. Empirical formula is the formula in which the simplest whole number ratio between numbers of atoms of each element is represented. The mass of empirical formula compound of any molecule is its empirical mass. The sum of each constituent atom’s molar mass multiplied by the number of atom of each element present in the compounds ...
Calculate the mass (in grams) of 2.125 moles of NaCl (molar mass - 58.44 g/mol), Give your answer to four significant figures. 9 Submit Answer Try Another Version 9 ltem attempts remaining Previous Next Mass 85.59 Determine the molar mass of potassium sulfite dihydrate, KSO3.2H2O (To avoid introducing rounding errors on intermediate calculations, enter your answer to five significant figures.)
Molarity of solution = 0.5\ M \text{Molarity} = \text{Moles of solute}/\text{Volume of solution in litres} Solute is \ \ \text{NaCl}. Molar mass of \ \ \text{NaCl} = 23 + 35.5 = 58.5\ g /mol \text{Number of moles} = \text{Given mass}/\text{Molar mass} Given mass of \ \ \text{NaCl} =5.85\ g Moles of \ \ \text{NaCl} = 5.85/58.5 =0.1 Volume of ...
M solute for the molar mass of the solute in g/mol. Therefore: Molarity = (msolute V soln × 100%) × 1000mL/L 100% ×M solute. Or, rewriting in terms of %w/v and implied unit cancellation, we have: Molarity(mol L) = 1000 Msolute %w/v 100%. As an example, if we have a 37% w/v aqueous HCl solution, then: Molarity(mol L) = 1000 36.4609 37 % w/v ...