You want 250 g of 35% H2SO4 by mass. That is 35 g/100 g. How much is that in 250 g? That will be 35g x (250g/100g) = 87.5 grams. (You can check that by 87.5/250 = 0.35 so that must be right.).
Now, you know you want 87.5 g, how much of the 92.5% stuff do you nee to obtain that.
100 g soln x (87.5g/92.5g) = about 94g. You would then want to add 250-94 = about 156 g water to make a total of 250 g. You need to do this more accurately. You can check out the final solution by 94.5g soln x (92.5/100) = approximately 87.5 g. Again, you should do this more accurately.
How many grams of a stock solution that is 92.5 percent H2SO4 by mass would be needed to make 250 grams of a 35.0 percent by mass solution?
Can you please teach me how to do this? I get the other stoichiometry stuff, but this one tripped me up because I don't know how to set up the problem.
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