I know that the idea is to find the final volume of water in the product solutio

Question

I know that the idea is to find the final volume of water in the product solution then subtract it from its initial volume in the feed solution which will give me the amount of water that was added to achieve the desired product solution. I spent more than an hour to figure out how to find the final volume of water in the product solution but I could not.
so please help me to solve this problem with clear and easy steps. I know that the idea is to find the final volume of water in the product solution then subtract it from its initial volume in the feed solution which will give me the amount of water that was added to achieve the desired product solution. I spent more than an hour to figure out how to find the final volume of water in the product solution but I could not.
so please help me to solve this problem with clear and easy steps. Run the proglan 8.95. An aqueous solution containing85 0 wt% H2SO4 at 60 F (specific gravity 1.78) is diluted with be ure liauid water at the same temperature. The feed solution volume is 350 mL. T (a) The productsolution is to contain 30.0 wt%H2SO4. Calculate the volume (mL) of water needed for the dilution,

Explanation / Answer

Feed solution volume = 350

Feed solution conc = 85% by wt

Mass of feed solution = Specific gravity*Volume = 350*1.78 = 623 g

So, Mass of water = 15% = 0.15*623 = 93.45 g

Mass of acid = 623-93.45 = 529.55 g

Assume that 'x' g of water is added

Total mass of water = 93.45+x

Now, % composition changes to 30%, which means:

529.55/(623+x) = 0.3

Solving we get:

x = 1142.2 g

So,

Volume of water needed = Mass/Density = 1142.2/1 = 1142.2 mL

Hope this helps !

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---