We have to know the molarity of solution obtained when 5.71 g of Na₂CO₃.10 H₂O is dissolved in water and made up to 250 cm³ solution.
The molarity of solution obtained when 5.71 g of sodium carbonate-10-water (Na₂CO₃.10 H₂O) is dissolved in water and made up to 250.0 cm^3 solutionis: (A) 0.08 mol dm⁻³
The molarit y of solution means the number of moles of solute present in one litre of solution. Here solute is Na₂CO₃.10 H₂O and solvent is water. Volume of solution is 250 cm³.
Molar mass of Na₂CO₃.10 H₂O is 286 grams which means mass of one mole of Na₂CO₃.10 H₂O is 286 grams.
5.71 grams of Na₂CO₃.10 H₂O is equal to 
= 0.0199 moles of Na₂CO₃.10 H₂O. So, 0.0199 moles of Na₂CO₃.10 H₂O present in 250 cm³ volume of solution.
Hence, number of moles of Na₂CO₃.10 H₂O present in one litre (equal to 1000 cm³) of solution is 
= 0.0796 moles. So, the molarity of the solution is 0.0796 mol/dm³ ≅ 0.08 mol/dm³
its 40
I just took the test right now to get this answer
The correct answer is 0.014467 M.
Molarity is defined as the number of moles present in a liter solution, that is, number of moles / liter solution.
The molar mass of sodium (Na) is 23.0 g/mol
Thus, 1 mole of Na contains 23.0 g
Now, x moles of Na contains 0.50 g
Moles = 0.50 × 1 / 23.0
Moles = 0.50 / 23.0
= 0.0217 moles of Na
Molarity = Number of moles / liters of solution
= 0.0217 / 1.5
= 0.014467 M
Answer:
the answer is b
Explanation:
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