Difficulty: Medium
Correct Answer: 90%
Explanation:
Introduction / Context:Fertilizer grade verification often uses nitrogen analysis. For urea, knowing the theoretical nitrogen fraction allows back-calculating the mass fraction of urea in a mixture or degraded sample. This is a routine quality-control calculation in fertilizer plants and agro-retail labs.
Given Data / Assumptions:
Concept / Approach:Theoretical nitrogen in pure urea is 28/60 by mass = 0.4667 → 46.67%. If a sample analyzes at 42% N, the mass fraction of urea is the ratio of measured N% to theoretical N%, assuming all nitrogen arises from urea.
Step-by-Step Solution:
Compute theoretical N% in pure urea: (28/60) * 100 = 46.67%.Given sample N% = 42.00%.Mass fraction of urea = 42.00 / 46.67 = 0.900 (approx.).Express as percent: 90% urea by mass.Verification / Alternative check:Take 100 g sample → contains 42 g N → equivalent urea mass = 42 / 0.4667 ≈ 90.0 g, confirming 90%.
Why Other Options Are Wrong:
Common Pitfalls:Using 50% as rough N content for urea (incorrect); ignoring that all nitrogen detected may not come from urea in adulterated samples (an assumption stated here).
Final Answer:90%
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