What is the molarity of Na+ion in a
solution prepared by dissolving
0.550 gram of Na2SO4 in a volume
of 100 ml water.
Answers
Answer:
molarity = no.of mol of solute/volume of solution in litres
= 0.550/142.04/0.1L
= 0.038 M
[Na+] = 0.076M
The molarity of the Na+ ion in the solution is 0.076 M.
Given: Mass of Na2SO4 = 0.550 g,
The volume of water = 100 ml
To Find: The molarity of Na+ ion in the solution.
Solution:
- We know that the molarity of a solution can be calculated using the formula,
Molarity (M) = number of moles of solute (n) / Volume ( in L )
- We know that the molarity of a compound is equal to the molarity of its constituent ions in the ratio of the molecular formula.
Hence, 1 mole of Na2SO4 contains= 2 moles of Na+ ion = 1 mole of S atom
Coming to the numerical, we are given that,
Mass of Na2SO4 = 0.550 g,
Molecular mass of Na2SO4 = ( 2×23 + 32 + 4×16 ) = 142 g
Thus, the number of moles of Na2SO4 = Mass of Na2SO4 /Molecular mass
= 0.55 / 142
= 3.873 × 10^-3
The volume of water = 100/1000 L
= 0.1 L
Hence, the molarity of Na2SO4 = number of moles of Na2SO4 /Volume
= 3.873 × 10^-3 / 0.1 M
= 0.038 M
Now, as said earlier,
1 mole of Na2SO4 contains= 2 moles of Na+ ion
the molarity of Na2SO4 = 0.038 M
Hence, the molarity of Na+ ions = 2 × molarity of Na2SO4
= 2 × 0.038 M
= 0.076 M
Hence, the molarity of the Na+ ion in the solution is 0.076 M.
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