What is the molarity of 0.25 dm^3 of a solution containing 10.35 g of k2CO3?
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Given:-
→Volume of the solution = 0.25dm³
→Mass of K₂CO₃ (solute) = 10.35g
To find:-
→Molarity of the solution
Solution:
•Atomic mass of Potassium(K) = 39u
•Atomic mass of Carbon(C) = 12u
•Atomic mass of Oxygen(O) = 16u
Thus, the molecular mass of K₂CO₃:-
= 39×2+12+16×3
= 78+12+48
= 138u
We know that molar mass is the molecular/atomic mass in grams. Thus, molar mass of K₂CO₃ is 138g.
Number of moles in 10.35g of K₂CO₃:-
= Given Mass/Molar mass
= 10.35/138
= 0.075 mole
Hence, 0.075 mole are there in 10.35g of K₂CO₃.
Now, let's convert the volume of the solution from dm³ to L.
=> 1dm³ = 1L
=> 0.25dm³ = 0.25×1
=> 0.25L
Molarity of a solution:-
= Moles of solute/Liters of solution
= 0.075/0.25
= 0.3 M
Thus, molarity of the solution is 0.3 M .
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