JEE Mains · Chemistry · STD 11 - 1. Some basic concept of chemistry
\(100\, \mathrm{~mL}\) of \(\mathrm{Na}_{3} \mathrm{PO}_{4}\) solution contains \(3.45\, \mathrm{~g}\) of sodium. The molarity of the solution is \(.....\times 10^{-2}\) \(\operatorname{mol} \,\mathrm{L}^{-1} \cdot(\) Nearest integer \()\) [Atomic Masses - \(\mathrm{Na}: 23.0\, \mathrm{u}, \mathrm{O}: 16.0\, \mathrm{u}, \mathrm{P}: 31.0 \,\mathrm{u}]\)
- A \(500\)
- B \(50\)
- C \(5\)
- D \(0.50\)
Answer & Solution
Correct Answer
(B) \(50\)
Step-by-step Solution
Detailed explanation
\(\mathrm{Na}_{3} \mathrm{PO}_{4} \longrightarrow \quad3 \mathrm{Na}\) \(\frac{1}{3} \times \frac{3.45}{23} \mathrm{~mol} \quad\) \(3.45 \mathrm{~g}\) \(\quad\quad\quad\quad\quad\quad\quad\frac{3.45}{23} \mathrm{~mol}\) therefore molarity of \(\mathrm{Na}_{3} \mathrm{PO}_{4}\)…
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