Question

# A sample of $$^{ 14 }{ { CO }_{ 2 } }$$ was to be mixed with ordinary $${ CO }_{ 2 }$$ for a biological tracer experiment. In order that $${ 10 }^{ 3 }{ cm }^{ 3 }$$ of the diluted gas at NTP should have $${ 10 }^{ 4 }$$ $${ dis }/{ min }$$, how many $$\mu Ci$$ of radiocarbon$$-14$$ are needed to prepare $$60 L$$ of the diluted gas?

Solution

## $$10 { cm }^{ 3 }$$ of the diluted gas at NTP $$={ 10 }^{ 4 }{ dis }/{ min }$$                                      $$=\dfrac { { 10 }^{ 4 } }{ 60 } dpsl$$$$\therefore 60 L\left( 60,000{ cm }^{ 3 } \right)$$ of the dilute gas at NTP $$=\dfrac { { 10 }^{ 4 }\times 60,000 }{ 60\times 10 } dps$$Thus, number of $$\mu Ci$$ of $$^{ 14 }{ { CO }_{ 2 } }$$ needed $$=\dfrac { { 10 }^{ 4 }\times 60,000 }{ 60\times 10\times 3.7\times { 10 }^{ 4 } } \left( 1\mu Ci=3.7\times { 10 }^{ 4 }dps \right)$$                                           $$=27.03\mu Ci$$ChemistryNCERTStandard XII

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