Question

1) calculate the mass of CuSo4 and 5H2o needed to make 100.0 ml of 0.35M CuSo4

2) why is it necessary to be careful in making up the standard solution.

3) if 8.09 ml of 0.350 M CuSo4 is mixed with 2.13 ml water, calculate the Cu2+ in the resulting solution

Answer 1

1) we know that

moles = molarity x volume (ml) / 1000

so

moles of CuS04.5H20 = 0.35 x 100 / 1000 = 0.035

now

mass = moles x molar mass

molar mass of CuS04.5H20 = 249.685 g/mol

so

mass of CuS04.5H20 = 0.035 x 249.685

mass of CuS04.5H20 = 8.74

so

8.74 grams of CuS04.5H20 is needed

2)

because it may give errors in you experiment
3)

we know that

for dilution

M1V1 = M2V2

so

0.35 x 8.09 = M2 x 10.22

M2 = 0.277

so

the molarity of CuS04 in the resulting solution is 0.277 M

now

CUS04 ---> Cu+2 + S042-

we can see that

[Cu+2] = [CuS04]

so

[Cu+2] = 0.277 M

so

the concentration of Cu+2 in the resulting solution is 0.277 M