Chlorine exists as ${ }^{35}\mathrm{Cl}$ $(\text{ atomic mass }=34.9688\mathrm{u};75.77\mathrm{\%})$ and ${ }^{37}\mathrm{Cl}\text{ (atomic mass }=36.9659\mathrm{u};24.23\mathrm{\%}\text{ ). }$ The atomic mass of nature existing chlorine is 

Given data:

• m₁ = 34.9688 u, f₁ = 75.77% = 0.7577
• m₂ = 36.9659 u, f₂ = 24.23% = 0.2423

Substitute the values:

Atomic Mass = (34.9688 × 0.7577) + (36.9659 × 0.2423)

Atomic Mass = 26.4957 + 8.9570 = 35.4527 u

Final Answer:

35.4527u