Question

#12

atm. pressure? a) 104.2 °C (b) 101.68 °CC ) 100.42 °Cd ) 101.26 °C 11. A diver has a breathing mixture of air that has a partial pressure of N2 of 1.44 atm. and O2 of 0.36 atm at a total pressure of 1.8 atm when he dives. Determine the solubility of nitrogen in his blood at this pressure, assuming his blood behaves like water. The Henry's Law constant for nitrogen in water is 6.1 x 10mol.L'atm"! a) 8.8 x 104 mol.L''b) 3.1 x 104 mol.L''c) 4.4 x 10-mol.L. d) 1.1 x 104 mol.L-' 12. The ocean contains 35.0 g of NaCl (F. Wt. = 58.5 g/mol) per 1000 g of water. What is the freezing point of ocean water, neglecting any other salts that may be dissolved in it? (Kr= 1.86 °C/m for water) a) 2.18 °Cb)) -1.1°C c) -10.8°Cd) -2.2°C

Answer 1

Ans 12 As we know- ATE= ikm) where, ATE = depression in freezing point i= Van't hoff factor K4 = depression constant m= molality Here we have mass of solute, Nace = 35.0g molar mass of solute, Nace = 58.5g/mol mass of solvent, water = 1000g molality = mass of solute x 1000 Molay mass of solute x mass of solvent (9) in 35.0 x1000 58.5 X 1000 = 0.598 m Also, i=2 [nace Nat +ce] Kf = 1.86°c/m plugging values, we get ATE= (2) (1.86) (0.598) » solvent - Tsolution = 2.225 oc - Tsolution = 2.225 Tsolution = 2. 2°c Hence, option (d) - 2.2°C Please is CORRECT Rate ...