Describe how you would prepare 2.44 kg of an aqueous solution that is 16.3% KOH by mass.
Describe how you would prepare 2.44 of an aqueous solution that is 16.3% by mass.
0.163 kg KOH. Add 0.163 kg KOH to 2.277 kg water.
2.44 kg solution x 0.163 = 0.40 kg KOH. Add 0.40 kg KOH to 2.84 kg water.
0.163 kg KOH. Add 0.163 kg KOH to 2.603 kg water.
2.44 kg solution x 0.163 = 0.40 kg KOH. Add 0.40 kg KOH to 2.04 kg water.
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Explanation:

The formula for mass percentage is:

The data are:

Substitute and solve:

Then, you will need to use 0.40kg of KOH and the amount of water needed to complete 2.44 kg of solution:

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.