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If the densit...

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If the density of a 3 M Sodium thiosulphate $(N a_{2} S_{2} O_{3})$ solution is 1.2 g/mL, then the molality of the solution is:

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Solution

3 M solution means 3 moles of sodium thiosulphate is present in 1 L of the solution.
Mass of ( $N a_{2} S_{2} O_{3}$ ) in solution = $3 \times m o l a r m a s s$
Molar mass of ( $N a_{2} S_{2} O_{3}$ ) = 158 g/mol
Mass of solute = 3 $\times$ 158 = 474 g
Since, Density = $\frac{m a s s}{v o l u m e}$
Mass of solution = 1.2 g/mL $\times$ 1000 mL = 1200 g
Mass of solvent = mass of solution - mass of solute
= 1200 - 474 = 726 g
Molality (m) = $\frac{number of moles of solute}{mass of solvent in g} \times 1000$
Molality = $\frac{3}{726} \times 1000$ = 4.13 m

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