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Is 2 percent NaCl hypotonic or hypertonoic?

Updated: 11/10/2020
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9y ago

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This is a hypertonic solution.

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Q: Is 2 percent NaCl hypotonic or hypertonoic?
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Related questions

Is a 2 percent salt solution hypertonic or hypotonic to a 4 percent salt solution?

hypotonic


How many grams of NaCl are in a 2 percent NaCl solution?

That refers to a mixture consisting of 2/100 of sodium chloride (salt) and 98/100 of something else (usually water).


A bag with a 2 percent solution is placed in a beaker that has a 10 percent solution is the solution in the bag hypotonic hypertonic or isotonic?

hypertonic


How do you prepare a solution 2 percent NaCl wv?

Dissolve 2g of NaCl in 100 cm3 water at normal temperature.


Does 2 percent NaCl cause crenation or hemolysis in red blood cells?

crenation


What is the osmolarity of 2 percent NaCl?

The osmolarity is 4 osmol/L.


What does the 2 in NaCl mean?

NaCl has not 2 in the formula unit.


How many grams of NaCl will be produced from 13g Cl2?

13g Cl2 * (1mol cl / 34.45 ) * (2 mol / 2 mol) * (58.44g NaCl / 1 mol NaCl) = 22g NaCl


What compounds in the liquid phase can be considered an electrolyte co 2 nacl h 2 o 2?

NaCl


What is an example of a hypotonic solution?

A hypotonic solution has less than normal tension: hypo = less, and tonic = tonicity, the concentration of solute. Examples of hypotonic solutions: (1) Sports drinks that contain salts / electrolytes (2) physiologically: a. 0.45% NaCl (half-normal saline solution); since normal saline is 0.9% NaCl, any solution less than 9% is hypotonic b. dextrose 2.5% in water c. dextrose 2% in water


The molecular weight of NaCl determined by studing freezing point depression of its 0.5 percent aqueous solution is 30. What the apparent degree of dissociation of NaCl is?

Theritical MM/ Observed MM = 1 + a (2-1) 58.5/30 = 1+a a = .95 or 95%


2 Predict and explain what systems are electrically conducting a solid NaCl b molten NaCl c An aqueous solution of NaCl?

molten NaCl and An aqueous solution of NaCl will be conducting due to the presence of free ions in these.