JEE Mains · Chemistry · STD 11 - 1. Some basic concept of chemistry
A sample of \(NaClO_3\) is converted by heat to \(NaCl\) with a loss of \(0.16\, g\) of oxygen. The residue is dissolved in water and precipitated as \(AgCl\). The mass of \(AgCl\) (in \(g\)) obtained will be (Given: Molar mass of \(AgCl = 143.5\, g\,mol^{-1}\) )
- A \(0.35\)
- B \(0.54\)
- C \(0.41\)
- D \(0.48\)
Answer & Solution
Correct Answer
(D) \(0.48\)
Step-by-step Solution
Detailed explanation
No of moles of oxygen in \(0.16\,g\) of oxygen molecule \( = \frac{{0.16\,g}}{{32\,g/mol}} = 0.005\,mole\) \(2NaCl{O_3}\xrightarrow{\Delta }2NaCl + 3{O_2}\) According to the reaction, \(3\) moles of \(O_2\) \(=\) \(2\) moles of \(NaCl\) \(=\) \(2\) moles of \(AgCl\) Molar mass…
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