JEE MAIN - Chemistry (2021 - 26th August Evening Shift - No. 18)
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]
[Atomic Masses - Na : 23.0 u, O : 16.0 u, P : 31.0 u]
Answer
50
Explanation
Molarity of Na3PO4 Solution = $${{{n_{N{a_3}P{O_4}}}} \over {volume\,of\,solution\,in\,L}}$$
$$ = {{{1 \over 3} \times {{3.45} \over {23}}mol} \over {0.1\,L}}$$
= 0.5 = 50 $$\times$$ 10$$-$$2
$$ = {{{1 \over 3} \times {{3.45} \over {23}}mol} \over {0.1\,L}}$$
= 0.5 = 50 $$\times$$ 10$$-$$2
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