Practice question
Question
In the Carius method, 0.25 g of a compound gave 0.574 g of AgBr . What is the percentage of bromine? (Atomic masses: Ag = 108, Br = 80)
Explanation
Given:
In the Carius method, 0.25 g of a compound gave 0.574 g of AgBr . What is the percentage of bromine? (Atomic masses: Ag = 108, Br = 80)
These values define the system as per NCERT data.
Formula:
Molar mass of AgBr = 108 + 80 = 188 g/mol.
This is the standard NCERT relation for this phenomenon.
Substitution & Calculation:
Mass of Br = 80 × 0.574/188 = 0.244 g. Percentage = 0.244 × 100/0.25 = 97.6% .
Result:
The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kgâ»Â¹ Kâ»Â¹, m/s², 10â»âµ are properly used as per NCERT.
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