100 mL of $\mathrm{Na}_{3} \mathrm{PO}_{4}$ solution contains 3.45 g of sodium. The molarity of the solution is $\_\_\_\_$ $\times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}$. (Nearest integer) [Atomic Masses - Na : $23.0 \mathrm{u}, \mathrm{O}: 16.0 \mathrm{u}, \mathrm{P}: 31.0 \mathrm{u}$]

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Some Basic Concepts of Chemistry

100 mL of

Na_{3} PO_{4}

solution contains 3.45 g of sodium. The molarity of the solution is

____

\times 1 0^{- 2} mol L^{- 1}

. (Nearest integer)
[Atomic Masses - Na :

23.0 u, O : 16.0 u, P : 31.0 u

]

Answer: 50

The sodium present in the solution is

n_{Na} = \frac{3.45}{23.0} = 0.15 mol

Each formula unit of

N a_{3} P O_{4}

supplies three sodium atoms, so the salt present is

0.15/3 = 0.05

mol, held in

100

= 0.100

L. The molarity is

M = \frac{0.05}{0.100} = 0.5 mol L^{- 1}

that is

50 \times 1 0^{- 2} mol L^{- 1}

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Integer answer