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The strength of an aqueous solutions of sodium nitrate is 0.425 g/25ml |
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Answer» Hello Dear. Answer ⇒ Option B. 0.2 M (or 0.2 mol/L) SOLUTIONS ⇒ GIVEN ⇒ Mass of the Sodium NITRATE = 0.425 grams. Volume of the Solution = 25 ml. = 25/1000 liter. = 0.025 liter. Molar Mass of Sodium Nitrate = 85 g/mol. [∵ 23 + 14 + 3 × 16 = 85] ∵ No. of moles of Sodium Nitrate(NaNO₃) = Mass/Molar Mass = 0.425/85 = 0.005 g Now, ∵ Molarity of the Solution = No. of moles of Sodium Nitrate ÷ Volume of the Solutions in liter. = 0.005/0.025 = 0.2 M. Thus, the Molarity of the Solution is 0.2 M. (or 0.2 mol/L). Hence, Option B. is Correct. Hope it helps. |
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