0,6 M of sodium chloride is 35,064 g.
To calculate the molarity (M) of the NaCl solution, first find the number of moles of NaCl. The molar mass of NaCl is approximately 58.44 g/mol, so 21.04 g of NaCl is about 0.360 moles (21.04 g ÷ 58.44 g/mol). Molarity is defined as moles of solute per liter of solution; therefore, for 750 ml (0.750 L), the molarity is 0.360 moles ÷ 0.750 L = 0.480 M. Thus, the molarity of the NaCl solution is 0.480 M.
To find the concentration in molarity (M), first calculate the number of moles of NaCl. The molar mass of NaCl is approximately 58.44 g/mol, so 34.6 grams of NaCl is about 0.592 moles (34.6 g / 58.44 g/mol). Then, divide the number of moles by the volume of the solution in liters: 0.592 moles / 1.5 L = 0.395 M. Therefore, the concentration is approximately 0.395 M.
m = moles/kg = 3.00/1.50 = 2.00 molal
The molarity of the solution is calculated by dividing the moles of solute (0.250 mol NaCl) by the liters of solution (2.25 L). Molarity = moles of solute / liters of solution Molarity = 0.250 mol / 2.25 L = 0.111 M
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.
5 M of NaCl is equivalent to 292,2 grams.
The concentration is the same !
A solution of NaCl 1 M.
i = isotonic molar [glucose] / isotonic molar [NaCl] i = 14 M / 7 M = 2 i = isotonic molar [glucose] / isotonic molar [NaCl] i = 14 M / 7 M = 2 i = isotonic molar [glucose] / isotonic molar [NaCl] i = 14 M / 7 M = 2 i = isotonic molar [glucose] / isotonic molar [NaCl] i = 14 M / 7 M = 2
To find the volume of 0.075 M NaCl solution that can be made, you can use the formula C1V1 = C2V2. Plugging in the values, we get (9.0 M)(450 mL) = (0.075 M)(V2). Solving for V2 gives V2 = 1.1 L. Therefore, 1.1 L of 0.075 M NaCl solution can be made by diluting 450 mL of 9.0 M NaCl.
5 % NaCl is equal to 5 g NaCl in 100 g of a material.
Pure water will have the lowest boiling point because it does not contain any solute particles to elevate the boiling point. As the concentration of NaCl increases, the boiling point also increases due to an increase in the number of solute particles that disrupt the formation of water vapor. Therefore, 0.5 M NaCl will have a higher boiling point than pure water, followed by 1.0 M NaCl, and finally 2.0 M NaCl will have the highest boiling point.
0.0002 mol NaCl/mLmolarity = (moles solute)/(L of solution)0.20 M NaCl = 0.20 mol NaCl/L1 L = 1000 mL0.20 mole NaCl/1000 mL = 0.00020 mol NaCl/mL (rounded to two significant figures)
To make 1 liter of 0.1 M NaCl solution, you will need 25 ml of the 4 M NaCl stock solution and 975 ml of water. This will give you the desired concentration of 0.1 M NaCl in a total volume of 1 liter.
Molarity = moles of solute/Liters of solutionGet moles NaCl.58.44 grams NaCl (1 mole NaCl/58.44 grams)= 1 mole NaCl------------------Molarity = 1 mole NaCl/1 liter= 1 M NaCl========
After the titration of 0.1 M HCl with 0.1 M NaOH, the material remaining in the flask would be sodium chloride (NaCl) along with water. The reaction between HCl and NaOH forms water and NaCl as the products.
To find the amount of NaCl in 1.84 L of 0.200 M NaCl, you use the formula: amount = concentration x volume. Thus, amount = 0.200 mol/L x 1.84 L = 0.368 mol of NaCl.