Sea water, which can be considered as a 6 molar (6 M ) solution of NaCl , has a density of 2 ~g ~mL ⁻¹ . The concentration of dissolved oxygen ( O ₂ ) in sea water is 5.8 ppm. Then the concentration of dissolved oxygen ( O ₂ ) in sea water, is x 10⁻⁴ ~m . x = _______. (Nearest integer) Given: Molar mass of NaCl is 58.5 ~g ~mol ⁻¹ Molar mass of O ₂ is 32 ~g ~mol ⁻¹

Step-by-step solution

Sea water is 6 Molar in NaCl , So 1000 ml of sea water contains 6 mol of NaCl. mass of solution = Volume density =1000 2 mass of solution =2000 ~g aligned & ppm = mass of O ₂ 2000 10^6 & mass of O ₂=5.8 2 10⁻³ & =1.16 10⁻² ~g aligned molality for O ₂= 1.16 10⁻² / 32 (2000-6 58.5) 1000 aligned & = 1.16 10 32 1649 = & 0.000219 = & 2.19 10⁻⁴ aligned Correct answer 2