How many mols of the stock solution do you want? That's M x L = 0.500 x 0.100 K2CrO4 = ?
For the 0.0025 M solution use the dilution formula.
mL1 x M1 = mL2 x M2
10 mL x 0.0025 = mL2 x 0.100M
Solve for mL2, pipet that volume and make to 10 mL.
Assuming the volumes add, the total volume will be 250 mL + 150 mL = 400 mL. So the FeSO4 has been diluted from 250 to 400 so the (FeSO4) = 0.350 x 250/400 = ?
(KCl) = 0.200M x 150/400 = ?
Describe how one would prepare these solutions in the laboratory, using weights reagents and volumes of solutions:
-500 mL of .100M K2Cr)4 stock solution from pure K2CrO4(s)
-10mL of .0025M K2CrO4(aq) from .100M K2CrO4(aq)
-What would the concentrations of FeSO4 and KCL be if 250mL of .350M FeSO4 were mixed with 150mL of .200M KCL? They do not react.
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