How many grams of a stock solution that is 87.0 percent H2SO4 by mass would be needed to make 200 grams of a 35.5 percent by mass solution? Show all of the work needed to solve this problem.

Question

anonymous · Answer

This is a typical dilution/concentration problem. To solve this you only need to know this equation$$c _{1}m _{1}=c _{2}m _{2}$$ where c1 stands for the initial concentration (this particular case 87%) and m1 stands for the initial mass (this particular case is our question)
$$m _{1}=(c _{2}m _{2})/c _{1}$$ $$m _{1}=(0.355*200)/.87$$ $$m1=81.609 grams$$