Question #f1e48

Question

1106 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-16T18:42:31

#25 g# of #Na_2SO_4# is required.

#C % = ("mass of solute") /("mass of solute"+"mass of solvent")#

Let, #rarr# mass of solute #(Na_2SO_4) = x#
#rarr# mass of solvent (water)#= 100g#

#C %= x/(x+100)*100%=20%#

Hence the equation becomes:

#x=(x+100)*0.2#

#x=0.2x+20#

#0.8x=20#

#x=20/0.8#

#x=25g#. (Answer).

Hope it Helps:)