The solution is an electrolyte
This is because NaCl is a strong electrolyte that dissociates completely into ions in water, leading to more solute particles that disrupt vapor pressure. In contrast, KNO3 is a weak electrolyte that partially dissociates, resulting in fewer solute particles and less disruption of vapor pressure.
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 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.
Given: 0.5 g NaCl; 0.05 L of solution.1) Find the molar mass of NaCl.(22.99) + (35.45) = 58.44 g/mol of NaCl2) Convert grams of NaCl to moles of NaCl.(0.5 g NaCl) X (1 mol NaCl / 58.44 g NaCl) = 0.00855578 mol NaCl3) Use the molarity equation, M = mol / L, to solve for molarity (M).(0.00855578 mol NaCl / 0.05 L) = 0.17 M
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.
A solution of NaCl 1 M.
This is because NaCl is a strong electrolyte that dissociates completely into ions in water, leading to more solute particles that disrupt vapor pressure. In contrast, KNO3 is a weak electrolyte that partially dissociates, resulting in fewer solute particles and less disruption of vapor pressure.
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========
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.
Glucose is not an electrolyte in solution.
5 M of NaCl is equivalent to 292,2 grams.
The concentration is the same !
To make 1 liter of 0.4 M NaCl solution, you would need 200 mL of 2 M NaCl solution. This can be calculated using the formula Mi x Vi = Mf x Vf, where Mi is the initial molarity, Vi is the initial volume, Mf is the final molarity, and Vf is the final volume.
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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 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.