10lts of water were added to ten lts of a 30% strong solution of sulphuric acid.what is the strength of resulting solution?

• In ten ltr solution,the quantity of sulphuric acid = .3*10=3lts
  now solution contain = 7+10 ltr water = 17ltr and 3lts of acid
  total solution =20lts
  soltuion is = 3/20 *100 = 15%
• 11 years agoHelpfull: Yes(32) No(2)
• sulphuric acid=SA and water=W
  so ratio in initial mixture SA:W=30:70
  so sulphuric acid =3 ltr and water= 7 ltr
  now in final mixture acid=3 ltr and water= 17 ltr
  ratio SA:W=3:17
  now strength={3/(3+17)}*100
  =15%
  (ANSWER)
• 11 years agoHelpfull: Yes(20) No(0)
• In ten ltr solution,the quantity of sulphuric acid = .3*10=3lts
  now solution contain = 7+10 ltr water = 17ltr
  soltuion is = 3/17 *100 = 17.647%
• 11 years agoHelpfull: Yes(2) No(14)
• quantity of sulphuric acid =30/100*10=3 lt
  now solution has 7 +10=17 lt water
  total solutin=20 lt
  strength =3/20*100=15%
  
• 11 years agoHelpfull: Yes(1) No(0)
• 10 ltrs of 30% strong solution of suslphuric acid has (30/100)*10=3 ltrs of water...when 10 ltrs of water is added to this solution,amount of water in resultant solution is 13 ltrs.so strength of solution is x....(x/100)*10=13=>x=130...strength of resultant solution is 130%
• 11 years agoHelpfull: Yes(0) No(16)