Question

If nickel crystallized into a body centered cubic unit cell, what density would it have (in g/cm') if it also had an edge length of 352 pm? Answer: Determine the AHvap of Anisole with a vapor pressure of 175.0 mmHg at 29°C whose normal boiling point is 154°C. Answer: You are given a solution of 80.0 g methyl salicylate, C7H602, dissolved in 400.0 g of benzene, C6H6. The Kf for benzene is 5.10°C/m and the freezing point is 5.50°C for benzene. Calculate the freezing point of the solution. Answer: You are given a solution of 80.0 g methyl salicylate, C7H602, dissolved in 400.0 g of benzene, C6H6. The Kb for benzene is 2.53°C/m and the boiling point is 80.1°C for benzene. Calculate the boiling point of the solution. Answer:

Answer 1

(1) Edge length a= 352pm as 352 pm 1cm 3.52x10 cm. 10 10pm NA = Avagadro's No.2 61022x183 atom mole for B cc No of atom per lattice 2 atem. = 2 Mw of Nickel = 58.7 groogmof 80 ZX Mw 3 density d = NAX a da 2 atom x 58.7 grmonof 6.022% 1623 atom X [3.527108cm] 3 mote > de 4.67 Gm cm3 2-) Pi 175 mmHg at T= 29°C= 302K PF latm=760 mm/ mm Hg at Telsuca 427 K (.. Boiling point is measured at tatom ) by clausis - clapeys. on equation 80 3) hm (760 DH I 302 K mol-k I 427K 8.314 I DH = 12595.6 I met

• 80.0gm Mass of methyl salicylate Mw of Cq Ho O = [7x12 + 6X1+ 2x 16] 122.9mmel 00 656 mole SO mole of CHO 80 m 122 goland Mass of solvent 400gmx 1kg looogm 80 0,656 mole 1.64 m 0.4 Kg 0.4kg 80 Dip - Kf " => sth (5.10°C/6) X 10649 DIf = 8.36°C 80 If = (5.50 50- 8.36 360°c = – 2.86 - 2.86°C DT 6= mx K b DT 6= (1.648 2.53) °c = 4.15°C 80 To= (80. 1+4.15) °C 20 Z - T6= 84.25°C

Please check the formula of methyl salicylate...

Because the formula for it is C8H8O3.

Here it is given as C7H6O2..

But i have solved as the given formula in the question..

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