The density of a `3 M Na_(2) S_(2) O_(3)` (sodium thiosulphate) solution is `1.25 g mL^(-1)`. Calculate:
a. % by weight of `Na_(2) S_(2) O_(3)`
b. Mole fraction of `Na_(2) S_(2) O_(3)`
c. Molalities of `Na^(o+)` and `S_(2) O_(3)^(2-)` ions.

Explanation

Calculation of % nass if `Na_(2)S_(2)O_(3)`
`"Mass of "Na_(2)S_(2)O_(3) "in 1M solution"=2xx23+2xx32+3xx16=158 g`
Mass of `Na_(2)S_(2)O_(3)` in 3M solution`=3xx158=474 g`
Volume of solution=100-0mL,Density of solution=1.25g/mL
`"Mass of solution" =(V xx d)=(1000mL)xx(1.25g/mL)=1250 g`
% mass of `Na_(2)S_(2)O_(3)`=474 g
Mas of water=1250-474-776 g
`" No.of moles of "Na_(2)S_(2)O_(3)=("Mass of" Na_(2)S_(2)O_(3))/(Molar mass)=((474 g))/((158 gmol^(-1)))=3 mol`
`"No. of moles of "H_(2)O=("Mass of water")/("Molar mass")=((776g))/((18g moil^(-1)))=43.11 mol`
`" Mole faction of "Na_(2)S_(2)O_(3)=("N0.of moles of" Na_(2)S_(2)O_(3))/("No. of moles of "Na_(2)S_(2)O_(3)+"N0.of moles of water")`
`=((3 mol))/((3 mol+43.11))=0.065`
Calculate of molality of molality of Na+ inos and `S_(2)O_(3)^(2-)`
Sodium thiosulpate ionises as:
`Na_(2)S_(2)O_(3)Leftrightarrow2Na^(+)+s_(2)O_(3)^(2-)`
`"No. of moles of "Na^(+)ions=2xx3 mol=6 mol`
`"No. of moles of" S_(2)O_(3)^(2-)"inos"=-1xx3 mol=3 mol`
` "Molality of "Na^(+)"inos"=("No.of moles of" Na^(+)"inos")/("Mass of water in kg")=((6mol))/((0.776 kg))=7.73 m`
`"Molality of "S(2)O_(3)^(2-)"inos"=("No.of moles of "S(2)O_(3)^(2-)"ions")/("Mass of water in kg")=((3mol))/((0.776 kg))=3.86 m`