Question

Question 25 of 26

Answer 1

Answer:

Total mass of leaked fluorine in six months= 0.405 kg

Explanation:

From the given,

Leaked mass of freon = 35.12 gm

Molecular mass of freon = $C_{2}H_{2}F_{3}Cl = (2\times 12)+(2\times1)+(3\times19)+(1\times35.5)=118.5g/mol$

Total weeks per month = 4

Fluorine released into the air in total weeks = $6/times4=24$

3 molecules of fluorine is present in the freon$C_{2}H_{2}F_{3}Cl$

${\tex Mass\,leak\,rate\,of\,fluorine}\,= \frac{Mass\,of\,fluorine\,in\,freon}{Molecularmass\,of\,freon}\times leak\,rate$

$\frac{19\times 3}{118.5}\times 35.12=16.8gm/week$

Total mass of fluorine leaked in six months = $24\times16.9=405.6gm= 0.405 kg$