Question

A stock concentration of 10000 ppm of jasmonic acid (MW : 210.27g/mol) was prepared in 50uL 80% methanol. What is the molar concentration?​

Answer 1

Answer:

=

106

10.6

×

V(ml)

1000

=

106

10.6

×

100

1000

=1M

Now, M

1

V

1

=M

2

V

2

1×10=M

2

×1000

M

2

=

1000

10

M

2

=10

−2

M

Answer 2

106

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