Correct Answer – B
`”Molecular weight of “Na_(2)S_(2)O_(3)=158g” mol”^(-1)`
From, molarity
`=(%”by weight of solute” xx”density of solution”xx10)/(M)`
`3=(%”by weight of”Na_(2)S_(2)O_(3)xx1.25xx10)/(158)`
`:.%”by weight of “Na_(2)S_(2)O_(2)=(3xx158)/(1.25xx10)=37.92%`

Correct Answer – B
Mass of 1000 mL of `Na_(2)S_(2)O_(3)` solution `=1.25xx1000=1250 g`
Molarity `=(“mass”)/(“molecular mass ” xx ” volume (mL)”)xx100`
`rArr 3=(“mass ” xx 1000)/(158xx1000)`
`therefore` Mass of `Na_(2)S_(2)O_(3)` is 1000 mL of 3M solution
`=3xx158=474 g`
Percentage by weight of `Na_(2)S_(2)O_(3)` in solution
`=(474)/(1250)xx100 =37.92 %`

(i) Mass of 1000 mL of `Na_(2)S_(2)O_(3)` solution
`=1.25 xx 1000 = 1250 g`
Mass of `Na_(2)S_(2)O_(3)` in 1000 mL of 3M solution
`= 3 xx` Mol. Mass of `Na_(2)S_(2)O_(3)`
`= 3xx 158 = 474g`
Mass percentage of `Na_(2)S_(2)O_(3)` in solution
`= (474)/(1250)xx100=37.92`
Alternatively, `M=(x xxd xx10)/(m_(A))`
`3=(x xx1.25xx10)/(158)`
`x = 37.92`
(ii) No. of moles of `Na_(2)S_(2)O_(3)=(474)/(158)=3`
Mass of water `=(1250-474)=776g`
No. of moles of water `= (776)/(18)=43.1`
Mole fraxtion of `Na_(2)S_(2)O_(3)=(3)/(43.1+3)=(3)/(46.1)=0.065`
(iii) No. of moles of `Na^(+)` ions
`=2xx` No. of moles of `Na_(2)S_(2)O_(3)`
`= 2xx 3=6`
Molality of `Na^(+)` ion `= (“No. of moles of “Na^(+)”ions”)/(“Mass of water in kg”)`
`=(6)/(776)xx1000`
`=7.73m`
No. of moles of `S_(2)O_(3)^(2-)` ions = No. of moles of `Na_(2)S_(2)O_(3)`
`=3`
Molality of `S_(2)O_(3)^(2-)` ions `= (3)/(776) xx 1000 = 3.86 m`.