Question

The streptomycin in $500\mathrm{g}$ of a broth was determined by addition of $1.25\mathrm{mg}$ of the pure antibiotic containing ${}^{14}\mathrm{C}$. The specific activity of this preparation was found to be $240\mathrm{cpm}/\mathrm{mg}$ for a 30 -min count. From the mixture, $0.112\mathrm{mg}$ of purified streptomycin was isolated, which produced 675 counts in $60.0\min$. Calculate the concentration in parts per million streptomycin in the sample.

Short answer

The concentration is$3.30ppm$streptomycin in the given sample.

Step-by-step solution

Given

Mass of broth$=500g$

Mass of tracer$=m_t=1.25mg$

Specific activity of tracer$=240cpm/mg$

Isolated mass$=m_m=0.112mg$

Count of mixture$=675countsin60.0min$

Calculation

Activity of the tracer can be calculated as:

$$\begin{gathered} R_t=specificactivity\times massoftracer \\ {R}_t=240\times 1.25 \\ {R}_t=300cpm \end{gathered}$$

The activity of mixture can be calculated as:

$$\begin{gathered} R_m=\frac{Count}{time} \\ {R}_m=\frac{675}{60} \\ {R}_m=11.25\mathrm{cpm} \end{gathered}$$

The amount of streptomycin is calculated by the isotope dilution method which is given as:

$m_x=\frac{R_t}{R_m}\times m_m-m_t$

Where,

$m_x=$mass of inactive sample

$R_t=$activity of tracer

$R_m=$activity of mixture

$m_m=$isolated mass which is purified from the mixture of radioactive tracer

$m_t=$mass of radioactive tracer

By substituting the given and calculated values in the above equation, we get,

$$\begin{gathered} m_x=\frac{300}{11.25}\times 0.112-1.25 \\ {m}_x=1.736mg \end{gathered}$$

Now, for converting in parts per million-

Amount of brothlocalid="1646459048262" $=500g$

Amount of streptomycin localid="1646459069150" $=1.736\mathrm{mg}$

localid="1646459188905" role="math" $$\begin{gathered} ppm=1.736\times \frac{{10}^3\text{microgram}}{mg}\times \frac{1}{500\mathrm{g}} \\ ppm=3.4ppm\approx 3.30ppm \end{gathered}$$

Final answer

The concentration in the given sample is calculated to be $3.30ppm$streptomycin.