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βˆ™ 15y ago

.500 M

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0.250 M

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Q: How many Na plus ions are in 250 mL of 0.500 M NaCl solution?
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Related questions

How many ions present in NaCl solution?

2 Sodium and Chloride.


How many ions are there in one mole of NaCl?

One mole of NaCl contains Avogadro's number of ions, which is approximately 6.022 x 10^23 ions. Each molecule of NaCl dissociates into one sodium ion (Na+) and one chloride ion (Cl-) in solution.


How many electrolytes does NaCl have?

In solution or melted sodium chloride is an electrolyte containing ions Na+ and Cl-.


Is salt in water an electrolyte solution?

Yes. All salts are electrolytes because it contains free ions ( like sodium and chlorin ions)... and also because it is an electrolyte solution. ---------------------------------------- Not all salts are electrolytes.


If a solution is acidic it has how many ions?

If a solution is basic it has how many ions


How many sodium ion and chloride ions are present in 30g of Nacl?

30 g NaCl = 0.513 mol NaCl So No. of Na+ ions = 3.088 X 1023 So No. of Cl- ions = 3.088 X 1023


How many percent chlorine found in human bodies?

Cholrine is present only 0.15% in human body.amount of chlorine and sodium is equal.which means chlorine is in form of Nacl .This Nacl may be in form of ions present in solution.


How many Na ions are in 4.2 g of NaCl?

2


How many sodium ions are in 2 moles of NaCl?

There are 1 mole of sodium ions in 1 mole of NaCl, as there is one sodium ion for each chlorine ion in the compound. Therefore, in 2 moles of NaCl, there are 2 moles of sodium ions, which is equal to 2 x 6.022 x 10^23 sodium ions.


How many grams of NaCl are in 225 grams of an 18 percent NaCl solution?

The answer is 8 g NaCl.


How many sodium ions are in one formula unit of NaCl?

There is 1 sodium ion (Na+) in one formula unit of NaCl.


How many moles of Na plus ions are there in a moles of NaCl?

There is one mole of Na⁺ ions for every mole of NaCl. This is because each mole of NaCl dissociates into one mole of Na⁺ ions and one mole of Cl⁻ ions in solution.