0.5 M
0.494 M
If we add 1 L of water to 1 L of 2 M NaCl solution will give 2 L of 1 M NaCl solution
molten NaCl and An aqueous solution of NaCl will be conducting due to the presence of free ions in these.
concentration or molarity = number of moles/volume number of moles (n) = mass in grams of nacl/relative atomic mass of nacl n=17.52/(23+35.5) n = 0.2994872 mol volume = 2000/1000 = 2dm^3 molarity = 0.2994872/2 =0.15mol/dm^3
2 Sodium and Chloride.
0.494 M
If we add 1 L of water to 1 L of 2 M NaCl solution will give 2 L of 1 M NaCl solution
molten NaCl and An aqueous solution of NaCl will be conducting due to the presence of free ions in these.
No, it is far from isotonic. there's even more salt in it than in ocean water (3%).A typical home recipe for an isotonic solution, used for nasal spray, consists of 1⁄2 litre of water, 4 to 5 grams (1 teaspoon) of salt. This is about 10 grams per Litre or 1 % NOT 10%
200 milliliters
That refers to a mixture consisting of 2/100 of sodium chloride (salt) and 98/100 of something else (usually water).
If concentration of Hydrogen in solution is 10-2 then its pH must be 2.
By the definition of molarity, which is mass of solute in moles divided by solution volume in liters, 250 ml of 0.15 M NaCl* solution requires (250/1000)(0.15) or 0.0375 moles of NaCl. Each liter of 2M NaCl solution contains 2 moles of NaCl. Therefore, an amount of 0.0375 moles of NaCl is contained in (0.0375/2) liters, or about 18.75 ml of the 2M NaCl, and if this volume of the more concentrated solution is diluted to a total volume of 250 ml, a 0.15 M solution will be obtained. _________________ *Note correct capitalization of the formula.
concentration or molarity = number of moles/volume number of moles (n) = mass in grams of nacl/relative atomic mass of nacl n=17.52/(23+35.5) n = 0.2994872 mol volume = 2000/1000 = 2dm^3 molarity = 0.2994872/2 =0.15mol/dm^3
The resulting product would be a roughly 2 M NaCl solution (slightly less than 2 molar because the solution is diluted by the water that is produced by the reaction). 2 M HCl + 1 M Na2CO3 --> 2 M NaCl + 1 M H2O + 1 M CO2 Two moles of NaCl weigh about 117 g (58.5 grams per mol) so the resulting solution has a sodium chloride concentration of about 117 grams per liter. This concentration is well below the maximum solubility of water at room temperature and atmospheric pressure ( > 300 g/liter) so there would not be any NaCl at all precipitating. The answer is thus: no NaCl precipitate will form.
This is a hypertonic solution.
Need moles NaCl first. 17.52 grams NaCl (1 mole NaCl/58.44 grams) = 0.29979 moles NaCl =====================Now. Molarity = moles of solute/Liters of solution ( 2000 ml = 2 Liters ) Molarity = 0.29979 mole NaCl/2 Liters = 0.1499 M NaCl ----------------------