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400g of 20%,240g of 43%and350g of 66% sulphuric acid are mixed .how much watershould be added to make a 10% solution of sulphuric acid?

Guest Dec 29, 2017
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If pure water is added, the  %  of   sulphuric acid   =   0%

 

So....we have the following equation :

 

400(.20)  + 240(.43)  + 350(.66)  +  0x =   .10(400 + 240 + 350  + x)

 

80  +  103.2  + 231    =   .10x  +  .10( 400 + 240  +  350)

 

414.2      =  .10x   +  .10(990)

 

414.2  =  .10x  +  99           subtract  99 from both sides

 

315.2  =  .10x        divide both sides by .10

 

3152  =  x

 

So.....3152g  of pure water should be added

 

 

cool cool cool

CPhill  Dec 29, 2017

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