M (NaCl) = 58.44 g/mol
M (SrCl2) = 158.53 g/mol
1 L of solution 1 contains 0.5 mol or (0.5 x 58.44 = 29.22 g) of NaCl
1 L of solution 2 contains 0.25 mol or (0.25 x 158.53 = 39.63) of SrCl2
29.22 g/L = 29220 micrograms / mL (NaCl)
39.63 g/L = 39630 micrograms / mL (SrCl2)
39630 > 29220
So solution 0.25 M SrCl2 has a larger concentration when expressed in microgram/mL
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