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You need to mention the unit for 2,61.

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Q: When 2.61 of solid Na2CO3 is dissolved is sufficient water to make 250 mL of solution the concentration of Na2CO3 is?
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What mass is required to make 2N solution of Na2co3 calculate mass of sodim carbonate which will be dissolved in 500 ccc?

50ml


What is the weight in grams of a 0.5 N Na2CO3 used in determining the concentration of unknown HCl solution?

A 0.5N Na2CO3 used in determining the concentration of an unknown HCl solution has a weight of 1.06 grams. To find the weight, you need to first find out how many moles there are by calculating molarity times volume.


Standardization of hcl solution with primary standard na2co3 solution?

solution containig (3.3)gm na2co3 .h2o in (15ml)


Sodium carbonate equation?

The chemical formula of sodium carbonate is Na2CO3. A solution hasn't a formula.


If 6.73g of Na2CO3 is dissolved in enough water to make 250ml of solution what is the molarity of the sodium carbonate what are the molar concentrations of the Na plus and CO32- ions?

6.73g (1 mole of Na2CO3/106g) = 0.063mol/0.25L = 0.25M Na2CO3 2 moles of Na+ x 0.25M = 0.5M Na+ 1 mole of CO32- x 0.25M = 0.25M CO3


What is the purpose of adding MnSO4 in Winkler's methode?

determine the dissolved oxygen(BOD).iodometryMnSO4+Na2CO3


What is the resulting concentration of Na When 70 milliliter of 3 molar Na2CO3 is added to 30 milliliters of 1 molar NaHCO3?

You have .07 Liters of 3 moles/Liter of Na2CO3. So if you do .07*3*2 (you multiply by two because there are TWO Na+ ions in Na2CO3) you get .42 moles of Na+. Then you do the same with NaHCO3. So, .03*1 is equal to .03 moles of Na+. Adding .42 with .03 will give you .45, the number of moles of Na+ of the whole solution. Since you are looking for concentration (which is moles if solute divided by Liters of solution), you must divide by .1 Liters (you get that by adding .07 and .03 of the two liquids that compose the solution) to get 4.5 Molar. That is the answer!


How do you prepare 250ml 15m solution of naco3 in a volumetric flask do presume a solvent ie h20 but then you have to presume volumes and conc and because 15M wanted you cant use the flask to dilute?

Did you mean 250 mL of 0.15 M Na2CO3? (It's impossible to make a 15 M Na2CO3 solution, as Na2CO3 is not that soluble.) Yes, you can assume that this will be an aqueous solution. Steps. 1. Calculate the mass of solid Na2CO3 needed. 2. Place this mass of Na2CO3 in the volumetric flask. 3. Add some water and swirl to dissolve the Na2CO3. 4. Carefully add more water until the total volume of solution is 250 mL, as indicated by the line etched on the neck of the volumetric flask. 250 mL x 1 L x 0.15 mol Na2CO3 x 105.99 g Na2CO3 = 4.0 g Na2CO3 needed ........... 1000 mL ......... 1 L ..................1 mol Na2CO3


How are 0.50 mol Na2CO3 and 0.50M Na2CO3 different?

0.5 mol of Na2CO2 is the ACTUAL dry crystal amount. 0.5 M 0f Na2CO3 means that o.5 moles has been dissolved in 1 litre of water. The M means ' moles per litre(dm^3)'.


What happens when you mixed the magnesium chloride solution with the sodium cabonate solution?

MgCl2 + Na2CO3 = 2NaCl + MgCO3


Why Na2CO3 is added to solution before Benedicts test is performed?

Sodium carbonate is added with the purpose to increase the pH of the solution.


What salt is produced in pH rainbow tube?

The solution must contain Na2CO3 and NaHCO3.