Question #d2535

1 Answer
Aug 17, 2017

#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#