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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 - \(\mathrm{Na}: 23.0\, \mathrm{u}, \mathrm{O}: 16.0\, \mathrm{u}, \mathrm{P}: 31.0 \,\mathrm{u}]\)

  1. A \(500\)
  2. B \(50\)
  3. C \(5\)
  4. D \(0.50\)
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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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