Answer to Question #69911 in General Chemistry for sonia

Question

If you are provided with a 200 ppm solution of quinine, how much would you need to transfer to a 50 mL volumetric flask to prepare a 500 ppb solution? What would its concentration be in mol L-1?

Answer 1

1 ppm = 1000 ppb
c_1×V_1=c_2×V_2
200×1000×V_1=500×0.05
V_1=(500×0.05)/(200×1000)=0.000125 (l)=0.125 (ml)
1 ppb = 1×〖10〗^(-6) g/l
500 ppb = 500×〖10〗^(-6) g/l
n(C_20 H_24 N_2 O_2 )=(500×〖10〗^(-6))/324,42=1.5×〖10〗^(-6) (mol)
c(C_20 H_24 N_2 O_2 )=1.5×〖10〗^(-6) (mol/l)
Answer: 0.125 ml; 1.5*10-6 mol/l.