A solution contains `1 mCi` of `L`-phenylalanine `C^(14)` labelled in `2.0 mL` solution. The specific activity of labelled sample is given as `150 mCi "mmol"^(-1)`. Calculate
(a) The concentration of the sample in the solution in `"mol" L^(-1)`
(b). The activity of solution in terms of counting per minute per `mL` at counting of 80%

Explanation

a. 1 mmol ` = 150 mCi`
`1 mCi = (1)/(150)` mmol `= (1)/(150 xx 2) = 3.33 xx 10^(-2) M`
b. `1 Ci = 3.7 xx 10^(10) dps`
`= 3.7 xx 10^(10) xx 60 dpm`
`= 3.7 xx 10^(10) xx 60 xx (80)/(100)` counts `"min"^(-1)`
`= (177.6 xx 10^(10))/(2 mL) = 88.8 xx 10^(10)` counts `"min"^(-1) mL^(-1)`
` mCi = 88.8 xx 10^(7)` count `"min"^(-1) mL^(-1)`