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  • 100 mL of solution contains 3.45 g of sodium. The molarity of the solution is ______<img alt="\mathrm{\times 10^{-2}\: mol\: L^{-1}

100 mL of \mathrm{Na_{3}PO_{4}} solution contains 3.45 g of sodium. The molarity of the solution is ______\mathrm{\times 10^{-2}\: mol\: L^{-1}.} (Nearest integer) [Atomic Masses - \mathrm{Na:23. 0u ,\ O:16. 0u ,\ P:31. 0u} ]
 

Answers (1)

best_answer

Given,

Mass of Na = 3.45 g
\mathrm{\therefore moles\: of\: Na= \frac{3. 45}{23}= 0. 15}
\begin{aligned} \mathrm{\therefore moles\: of\: Na_{3}PO_{4}}&= \mathrm{\frac{moles\, of\, Na}{3}} \\&=\frac{0.15}{3} \\&=0.05 \end{aligned}
\therefore 100 ml contains 0.05 moles

\therefore 1 lit contains 0.5 moles

Thus, Molarity =0.5= 50\times 10^{-2}

Hence, the correct answer is 50

Posted by

sudhir.kumar

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