100 mL of Na3PO4 solution contains 3.45 g of sodium. The molarity of the solution is _____________ $\times$ 10$-$2 mol L$-$1. (Nearest integer)
[Atomic Masses - Na : 23.0 u, O : 16.0 u, P : 31.0 u]
Answer (integer)
50
Solution
Molarity of Na<sub>3</sub>PO<sub>4</sub> Solution = ${{{n_{N{a_3}P{O_4}}}} \over {volume\,of\,solution\,in\,L}}$<br><br>$= {{{1 \over 3} \times {{3.45} \over {23}}mol} \over {0.1\,L}}$<br><br>= 0.5 = 50 $\times$ 10<sup>$-$2</sup>
About this question
Subject: Chemistry · Chapter: Some Basic Concepts of Chemistry · Topic: Mole Concept
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