Question

Question #d2535

Answer 1

`1 mol L^-1`

Explanation:

Firstly recall that;

`1000ml = 1L`

`(Na = 23, color(white)(x) O = 16, color(white)(x) S = 32)`

`Molarity = "Moles of solute"/"Volume of Solvent or Solution"`

`"Moles of solute" = "Mass of solute"/"Molar mass of solute"`

`"Volume of solvent" = 1000ml = 1L`

`"Solute" = Na_2SO_4`

`"Mass of solute" = 142g`

`"Molar mass of solute" = (23 xx 2) + 32 + (16 xx 4) gmol^-1`

`color (white) (XXXXXXXXXX) = 46 + 32 + 64 gmol^-1`

`color (white) (XXXXXXXXXX) = 142 gmol^-1`

`rArr "Moles of solute" = (142cancelg)/(142 cancelgmol^-1)`

`rArr "Moles of solute" = 1mol`

`rArr Molarity = "Moles of solute"/"Volume of Solvent or Solution"`

`rArr Molarity = (1 mol)/(1 L)`

`rArr Molarity = 1 mol L^-1`