MCQ
$100\, \mathrm{~mL}$ 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
    $500$
  • $50$
  • C
    $5$
  • D
    $0.50$

Answer

Correct option: B.
$50$
b
$\mathrm{Na}_{3} \mathrm{PO}_{4} \longrightarrow \quad3 \mathrm{Na}$

$\frac{1}{3} \times \frac{3.45}{23} \mathrm{~mol} \quad$ $3.45 \mathrm{~g}$

$\quad\quad\quad\quad\quad\quad\quad\frac{3.45}{23} \mathrm{~mol}$

therefore molarity of $\mathrm{Na}_{3} \mathrm{PO}_{4}$ Solution =

$\frac{\mathrm{n}_{\mathrm{Na}_{3} \mathrm{PO}_{4}}}{\text { volume of solution in } \mathrm{L}}$

$=\frac{\frac{1}{3} \times \frac{3.45}{23}\, \mathrm{~mol}}{0.1\, \mathrm{~L}}$

$=0.5=50\, \times 10^{-2}$

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