Question #b312e

Question

1047 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-11T09:22:54

1st part

#"Molar mass of " H_2SO_4 "=98g/"mol" #
#"Strngth of "H_2SO_4" solution""=7.5M#
#=7.5"mol"/Lxx98g/"mol"= 735g/L#

#"Volume of acid soln containing " 150g H_2SO_4=(150g)/(735g/L)=0.2041L=204.1mL#

2nd Part
#V_1->"Initial volume of HCl solution"=?#

#S_1->"Initial Strngth of HCl solution"=2.5M#

After dilution

#V_2->"Final volume of HCl solution"=350mL#

#S_2->"Fimal Strength of HCl solution"= 1.15M#

So

#V_1S_1=V_2xxS_2=>V_1xx2.5=350xx1.15#

#V_1=(350xx1.15)/2.5=161mL#