Express the concentration of 5% mass of solution of `Na_(2)S_(2)O_(3)` in terms of (a) molarity (b) molality and © mole fraction of solution. The density of solution at `25^(@)C` is gm `L^(-1)`

Explanation

Let the solution contain 5.0g of `Na_(2)S_(2)O_(3)` and 95.0 g of water. The number of moles of the two constitution are:
` "No. of moles of "Na_(2)S_(2)O_(3)=("Mass of" Na_(2)S_(2)O_(3))/("Molar mass") =((5.0g))/((158.0g mol^(-1))=0.0316 mol`
`"No. of moles of water"=("Mas of water")/("Molar mass")=((95.0g))/((18.0g mol^(-1)))=5.278 mol`
Volume of solution = `("Mass of solution")/("Deensity")=((100.0g))/((1.04 g mL^(-1)))=96.15 mL`
Calculation of molarity of solution
`"Molality of solution (m)"="(Moles of "(Na_(2)S_(2)O_(3))/("Volume of solution in litres")`
` =((0.0316 mol))/((0.095 kg))=0.333 mol kg^(-1)=0.333 m` .
Calculation of mole fraction of solute
Mole fraction of solute in solution may calculated as:
`"Mole fraction of solute "(x_(B)=n_(B))/(n_(B)+n_(A))=((0.0316 mol))/((0.0316 mol+5.278 mol))`
`=((0.0316 mol))/((5.3096 mol))=5.95xx10^(-3)`