A chemist dissolves 0.098 g of CuSO4 · 5H2O?
1a. A chemist dissolves 0.098 g of CuSO4 · 5H2O
in water and dilutes the solution to the mark
in a 500-mL volumetric flask. A 27.8-mL
sample of this solution is then transferred to a
second 500-mL volumetric flask and diluted.
What is the molarity of CuSO4 in the second
solution?
Answer in units of M
1b. To prepare the second 500 mL of solution
directly, what mass of CuSO4 · 5H2O would
need to be weighed out?
Answer in units of mg
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Verified answer
first calculate the molarity at start
concentration = 0.098 x 2g g per L
Molar mass of CuSO4*5H2O is 249.6861 ± 0.0003 g/mol .. let's call it 249.7
molarity = (0.098 x 2)/ 249.7 = 7.85 x 10^-4 M
so 1000mL contains 7.85 x 10^-4 moles
27.8mL contains 2.182 x 10^-5 moles
this is made up to 500mL
so in 1L there is 2.182 x 10^-5 x 2 = 4.36 x 10^-5 moles or 4.36 x 10^-5M
double check my actual fifures but you should now be able to understand this
for 1b ....mass 2.182 x 10^-5 x 249.7g