The density of $3$ M solution of sodium thiosulphate $(N a_{2} S_{2} O_{3})$ is $1.58$ g/ml. Calculate the molality of $N a^{+}$ and $S_{2} O_{3}^{2 -}$ ions.

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Solution

$H e r e M o l a r i t y o f s o l u t i o n i s 3 M d e n s i t y o f s o l u t i o n i s 1.589 / m l w e k n o w t h a t \frac{1}{m} = \frac{d}{M} - \frac{M_{s}}{1000} H e r e M_{s} = M a s s o f s o l u t e f o r N_{s} + M_{s} = 23 S o, \frac{1}{m} = \frac{1.58}{3} - \frac{23}{1000} \frac{1}{m} = 0.526 - 0.023 \frac{1}{m} = 0.503 m = \frac{1}{0.5.3} = {1.9880}_{m} F o r O_{3}^{- 2} M_{s} = 112 S o, \frac{1}{m} = \frac{1.58}{3} - \frac{112}{1000} \frac{1}{m} = 0.526 - 0.112 m = \frac{1}{0.414} = 2.415 m$

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Similar questions

Q. The density of

3

M solution of sodium thiosulphate

(N a_{2} S_{2} O_{3})

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g/ml., Calculate mole fraction of

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Q. The density of a 3 M sodium thiosulphate solution

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The density of 3M aqueous solution of sodium thiosulphate is 1.25g/ml. Calculate (1) mole fraction of sodium thiosulphte (2) molalities of Na^+ and S2O3^-2 ions.

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^{- 1}

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The density of 3M solution of sodium thiosulphate (Na2S2O3) is 1.58 g/ml. Calculate the molality of Na+ and S2O2−3 ions.

HereMolarityofsolutionis3Mdensityofsolutionis1.589/mlweknowthat1m=dM−Ms1000HereMs=MassofsoluteforNs+Ms=23So,1m=1.583−2310001m=0.526−0.0231m=0.503m=10.5.3=1.9880mForO−23Ms=112So,1m=1.583−11210001m=0.526−0.112m=10.414=2.415m

The density of $3$ M solution of sodium thiosulphate $(N a_{2} S_{2} O_{3})$ is $1.58$ g/ml. Calculate the molality of $N a^{+}$ and $S_{2} O_{3}^{2 -}$ ions.