0
What else can I help you with?
The concertration of a water solution of NaCl is 2.48m and it contains 806 g of water How much NaCl is in the solution?
To determine the amount of NaCl in the solution, you first need to calculate the moles of NaCl present. Using the given molarity (2.48 M) and the volume of the solution (assumed to be 806 g = 806 ml for water), you can find the moles of NaCl. Then, you convert the moles of NaCl to grams using the molar mass of NaCl (58.44 g/mol) to find the amount of NaCl in the solution.
Determine how much of each of the following NaCl solutions in grams contains 1.5g of NaCl 0.058 percent NaCl by mass?
In order to determine this, it is necessary to know what solution we are looking at. One we know that we can look at the grams in a mole of the substance and determine the percentages based on molecular weight.
How much gNacl and how much gwater needed for ready of 500g solution of NaCl 0.95 percent?
0.95% * 500 g = 4.75 g NaCl
You want 1 liter of 0.1 M NaCl and you have 4 M stock solution How much of the 4 M solution and how much dH2O will you measure out for this dilution?
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.
How much 2M Nacl solution would you need to make 250ml of 0.15M Nacl solution?
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.
How much nacl is required for a 200 mM NaCl solution?
You need 58,44 mg of ultrapure NaCl; dissolve in demineralized water, at 20 0C, in a thermostat, using a class A volumetric flask of 1 L.
How much 0.075 M NaCl solution can be made by diluting 450 mL of 9.0 M NaCl?
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.
How much NaCl is required for 100 mM NaCl solution?
To prepare a 100 mM NaCl solution, you would need to calculate the molecular weight of NaCl, which is approximately 58.44 g/mol (sodium's atomic weight is 22.99 g/mol and chlorine's is 35.45 g/mol). To make a 100 mM solution, you would need 0.1 moles of NaCl per liter of solution. This would be equivalent to 5.844 grams of NaCl per liter of solution.
How much sodium chloride needed to make 10ppm sodium solution?
2.5 g of Nacl is to be dissolve in 100ml of water gives 10ppm of Na solution.
What is the molarity of sodium sulphate in a solution prepared by dissolving 15.5g in enough water to make 35ml of solution?
Molarity = moles of solute/volume of solution Find moles NaCl 55 grams NaCl (1mol NaCl/58.44 grams) = 0.941 moles NaCl Molarity = 0.941 moles NaCl/35 Liters = 0.027 Molarity NaCl ( sounds reasonable as 55 grams is not much in 35 Liters of water, which would be about 17.5 2 liter sodas )
Will elevation in boiling point be same for 0.1 m NaCl and 0.1m sucrose solution?
No, the elevation in boiling point will not be the same for a 0.1 m NaCl solution and a 0.1 m sucrose solution. This is because the elevation in boiling point is directly proportional to the number of particles in the solution, known as the van't Hoff factor. NaCl dissociates into two ions (Na+ and Cl-) in solution, so it has a van't Hoff factor of 2, while sucrose does not dissociate and has a van't Hoff factor of 1. Therefore, the NaCl solution will have a greater elevation in boiling point compared to the sucrose solution.
How much would the freezing point of the water decrease if 4 mol of NaCl were added to 1 kg of water for water and i 2 for NaCl?
The freezing point depression can be calculated using the formula: ΔTf = i * Kf * m, where i is the van't Hoff factor (number of particles), Kf is the cryoscopic constant, and m is the molality of the solution. Given that i for NaCl is 2, and assuming Kf for water is 1.86 °C/kg/mol, the ΔTf for a 4 mol NaCl solution in 1 kg of water can be calculated.
