The strength of an aqueous solution of sodium nitrate is 0.425g/25ml . the molarity of this solution is
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Given ⇒
Mass of the Sodium Nitrate = 0.425 g.
Molar Mass of the NaNO₃ = 85 g/mole.
[∵ 23 + 14 + 3 × 16 = 85]
Volume of the solution = 25 ml.
= 25/1000 liter.
= 0.025 liter.
Now, Using the Formula,
No. of moles of Sodium Nitrate = Mass/Molar Mass.
= 0.425/85
= 0.005 moles.
∵ Molarity = No. of moles of Sodium Nitrate/Volume of the solutions in liter.
= 0.005/0.025
= 0.2 M.
Hence, the Molarity of the solution is 0.2 M.
Hope it helps.
Given ⇒
Mass of the Sodium Nitrate = 0.425 g.
Molar Mass of the NaNO₃ = 85 g/mole.
[∵ 23 + 14 + 3 × 16 = 85]
Volume of the solution = 25 ml.
= 25/1000 liter.
= 0.025 liter.
Now, Using the Formula,
No. of moles of Sodium Nitrate = Mass/Molar Mass.
= 0.425/85
= 0.005 moles.
∵ Molarity = No. of moles of Sodium Nitrate/Volume of the solutions in liter.
= 0.005/0.025
= 0.2 M.
Hence, the Molarity of the solution is 0.2 M.
Hope it helps.
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