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How much NaCl require for 0.15M nacl solution?

Anonymous

∙ 15y ago

Updated: 10/6/2023

To get 0.15 moles of NaCl,

mass of NaCl = moles * molar mass

= 0.15 * (23+35.5)

= 8.775 g

and 0.15M is 0.15 mole of NaCl in on liter solvent which equals 8.775 g then ypu add solvent up to 1 liter.

Wiki User

∙ 15y ago

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More answers

0,15 M NaCl equal a concentration of 8,776 g/L.

Wiki User

∙ 11y ago

The answer is 4,383 g.

Wiki User

∙ 11y ago

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Related Questions

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?

To determine the grams of NaCl in 0.058% NaCl solution, you first convert 0.058% to a decimal by dividing by 100 (0.00058). Then, you set up a proportion to find out how much of the solution contains 1.5g of NaCl. 1.5g of NaCl is equivalent to 0.00058x, where x is the total grams of solution. Solve for x to find how much of the solution is needed.

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?

To prepare 250 ml of 0.15M NaCl solution, you would need 10 ml of 2M NaCl solution. This can be calculated using the formula: (C1V1 = C2V2), where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.

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?

To find the molarity of the solution, first calculate the number of moles of sodium sulfate by dividing the mass (15.5g) by the molar mass of sodium sulfate (142.04 g/mol). Next, convert the volume of the solution to liters (35 ml = 0.035 L). Finally, divide the number of moles by the volume in liters to get the molarity of sodium sulfate in the solution.

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.

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