. a How many atoms are there in a metallic face-centered cubic unit cell b. What

Question

. a How many atoms are there in a metallic face-centered cubic unit cell b. What fracti ion of an atom belongs to a unit cell if the atom is found in the edge of the unit cell? 10. Arrange the following in increasing order of boiling point (lowest b.p. first): C. 0.30 m MgCl2 A. 0.20 m KNO D. 0.25 m CH:OH B. 0.80 m CCI E. 0.23 m Cas(PO4)2 11. A 250 ml solution containing 1.045 g of an enzyme exhibits an osmotic pressure of 1.50 torr at 27 °C. Determine molar mass of the enzyme. 12. What is the normal boiling point of 4.60 m aqueous solution of Fe(NO3)3 solution? (ks of water= 0.51 °C/m) 13. Chromium crystallizes in body center cubic crystal with 4.80 A on an edge. What is the density of Chromium in g/cm3? (1 A 108 cm). n nf ornanic comnound was dissolved in 54.60 g of benzene.

Explanation / Answer

9)

a)

From the diagram of the fcc structure there are 4 atoms in unit cell.

b)

If we found the atom at the edge of the unit cell then it is shared by four cubes.

so fraction of atom shared each cube = 1/4 th of atom

10)

Boiling point of any substance is directly proportional to the number of moles of solute particles present in the solution.

KNO3 -----> K+ + NO3-

so there are 0.2 m of K+ ions and 0.2 m of NO3- ions.

Hence total number of particles = 0.4 m particles

CCl4 ia molecular compound so there are only 0.8 m of CCl4 molecules.

MgCl2 ----> Mg+2 + 2Cl-

so there are 0.3 m of Mg+2 ions and 2(0.3)m of Cl- ions

Hence the number of particles = 0.3+0.6 = 0.9 m particles

CH3OH is a molecular compound so there are only 0.25 m of CH3OH molecules are present

Ca3(PO4)2 -------> 3 Ca+2 + 2 PO4 3-

so ther are 0.23 * 3 = 0.69 m of Ca+2 ions and 0.23*2 = 0.46 m of PO4 3- ions.

Hence total number of particles = 0.69 + 0.46 = 1.15m particles

So among all Ca3(PO4)2 has more number of solute paticles and it has more boiling point.

11)

Osmotic pressure = C*S*T

C = concentration of the compound = mass of solute/molar mass of solute * 1000/volume of solution = 1.045/M*1000/250

S = solution constant = 0.083

T = absolute temperature = 27+273 = 300

from the formula,

1.5/760 = 1.045/M*1000/250 * 0.083 * 300

M = molecular mass = 52734.88 g/mol

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