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Question

# The density of 3M solution of Na2S2O3 is 1.25 g mL−1. Calculate the % by weight of Na2S2O3 and molality of Na+ ion

A
62.08 %, 7.73 M
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B
37.92 %, 7.73 M
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C
42.73 %, 3.86 M
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D
58.58 %, 3.86 M
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Solution

## The correct option is B 37.92 %, 7.73 MGiven Na2S2O3 has molarity = 3 mol L−1. This implies that 3 mol of Na2S2O3 are present in 1 L of solution. ∴ Weight of Na2S2O3=Molar mass×Mole =3×158=474 g and Volume of solution = 1 litre =1000 mL Also density of solution is 1.25 gml−1 Hence weight of solution =1000×1.25=1250 g Weight of solvent i.e water =1250−474=776 g % by weight of Na2S2O3=Weight of Na2S2O3Weight of Na2S2O3+Weight of H2O ⇒4741250×100=37.92 3 mol of Na2S2O3 gives 6 mol of Na+ and 3 mol of S2O2−3 Molality of Na+=Mole of Na+Weight of water (g)×1000=6×1000776=7.732 m

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