0.5 M
The concentration of the NaCl solution is 29.25 g/L. This is calculated by dividing the mass of NaCl (58.5 g) by the volume of the solution (2 L).
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 concentration of a solution in grams per liter, you need to divide the mass of the solute (in this case, 80 grams of NaCl) by the volume of the solution (2 liters). Therefore, the concentration of the solution would be 80 grams divided by 2 liters, which equals 40 grams per liter.
To make a 1.00M NaCl solution from a 2.00M solution, you can dilute the 2.00M solution by adding an equal volume of solvent (like water). For example, mix 1 cm^3 of the 2.00M solution with 1 cm^3 of water to create a 1.00M solution.
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 concentration of the NaCl solution is 29.25 g/L. This is calculated by dividing the mass of NaCl (58.5 g) by the volume of the solution (2 L).
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
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.
To find the concentration of a solution in grams per liter, you need to divide the mass of the solute (in this case, 80 grams of NaCl) by the volume of the solution (2 liters). Therefore, the concentration of the solution would be 80 grams divided by 2 liters, which equals 40 grams per liter.
To calculate the concentration in ppm, first determine the molarity of the solution using the given mass of NaCl and volume of solution. Next, convert the molarity to ppm by multiplying by the molar mass of NaCl and 1,000,000 (since 1 ppm = 1 mg/L). Finally, adjust for the final volume to get the concentration in ppm.
That refers to a mixture consisting of 2/100 of sodium chloride (salt) and 98/100 of something else (usually water).
To prepare a 2% NaCl (w/v) solution, you would dissolve 2 grams of NaCl in enough water to make 100 mL of solution. This means you would add 2 grams of NaCl to a flask and then add water until the total volume reaches 100 mL.
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 ----------------------
To make a 1.00M NaCl solution from a 2.00M solution, you can dilute the 2.00M solution by adding an equal volume of solvent (like water). For example, mix 1 cm^3 of the 2.00M solution with 1 cm^3 of water to create a 1.00M solution.
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%
This is a hypertonic solution.
"2 times concentration" typically refers to a solution that has double the concentration of a standard or reference solution. For example, if a standard solution has a concentration of 1 M (molar), a 2 times concentration would be 2 M. This means there are twice as many solute particles per unit volume compared to the standard solution, which can impact the solution's properties and reactions.