Question

problem help 1-4!

TULSI Lumpletes the statement or answers the question. 283.15 lubility of carbon dioxide in water at 10.0 °C and a pressure of 3.00 atm. 1) lity of carbon dioxide is 0.0791 mol/L water at 10.0 °C and 1.00 atm. T=MRT - I -M 1) Calculate the solubility of carb The solubility of carbon di A) 12.6 M B) 0.0791 M → C) 0.0264 M D) 0.237 M E) 4.21 M 33=(1) T RTI 2) 2) Which of the following is true about cubic unit cells of crystalline metals? A) A simple cubic unit cell contains 1.66 x 10-24 moles of atoms. B) A body-centered cubic unit cell contains 3.32 x 10-24 moles of atoms. C) A face-centered cubic unit cell contains 6.64 x 10-24 moles of atoms. D) A simple cubic unit cell contains 1 atom. E) A body-centered cubic unit cell contains 2 atoms. F) A face-centered cubic unit cell contains 4 atoms. Answers A-C are correct, while D-F are not. H) Answers D-F are correct, while A-C are not. D'Answers A-F are correct. 92.906 moss 3) Niobium (Nb) is a metal that crystalizes with cubic unit cell that is body-centered (BCC). Nb has an atomic radius of 143 pm. What is its density in g/cm3 A) 9.33 g/cm3 B) 8.57 g/cm3 C) 17.1 g/cm3 6.000 x 100 alpoms imol. D) 4.28 g/cm3 E) 52.8 g/cm3 330 pm mm 1100cm 3) a atoms/ imol 92.9069 - 3. 3x . 10 from the cromozes. What kon 4) 4) An aqueous solution has a mole fraction of solute of 0.0385. What is the molality of this solution? mz amoh solute A) 0.0860 m B) 2.054 m C) 55.50 m D) 2.137 m E) 2.222 m kg solvent

Answer 1

Question 1 According to Henry's law, S → Solubiüty pt pressure of gas Ku Henry's const 2= KHcos PCO2 Given values a Scoz = 0.0191 mol/l PCO2 = 1 am = Scoz & PCoz 0.0191 - 0.0791 mol/lamm So at pressure 3 am. Solubility Sco2 = kHco PCo2 = 0.0791 x3 = 0.237 M

Question 2 - Option H for simple cubic unit cell 8 corneas x per comer alom a 1 atom for Bcc sao) na spore in 8 comer x I per comer lahm I body conhe atom a latom bis be Toal. Zatom for fcc ; 5811 8 comer x I per comer a lahom 6 face centered & I atomper - 1 x6 =3 atom unitcell Tor 4 atom

Question 3. 0 fra density of unit cell ad-ZM NA. 23 d density z no. of atom per unit cell NA → avogadro no. (6.022x1023 moll) a edge rength. M molar mass For Bcc z = 2 Relation b/w radius and edge length is Hy > a. il a Hr ng re radius of Nb - 4 x 143 pm 03 23 23 Volume = -23 3 3.6 x 10 cm

substituting an egn do zM NA 23 - 23 3 • 2 x 92,9 4 mol G.022x10 molx 3.6x10 cm - 8.579/cm?

Question 4 Mole fraction of solute of Solution of solvent & 0.0385 I * I - 0.0385 = 0.9615 Molality (m) a No. of moles of solube A Mass of solvent in ka 18 Molar mas of Hoa maus of Solvent (H20) - e. no. of moles x Molan bolecular mass 0.9615x18 17.3079 Molality (m) - 0.0385 x 1000 kg 17.307 2.222m