Answer to Question #127014 in General Chemistry for Rosé A

Question

1. Calculate the (a) molality (b) molarity and (c) molefraction of KI if the density of 20% (mass/mass) aqueous KI solution is 1.202gmL-1. (5)

2. A compound contains 42.3913% K, 15.2173% Fe, 19.5652% C and 22.8260% N. The molecular mass of the compound is 368u. Find the molecular formula of the compound. (5)

Answer 1

1 a) Molar mass of KI = 39 + 127 = 166 g mol^-1

20% (mass/mass) aqueous solution of KI means 20 g of KI is present in 100 g of solution.

That is,

20 g of KI is present in (100 - 20) g of water = 80 g of water

Thus, molality of the solution

 = moles of KI/mass of water in Kg = (20/166)/ 0.08

= 1.506 m

= 1.51 m

b) It is given that the density of the solution = 1.202 g mL^?1

mass/ density =Volume of 100 g solution

 100g/ 1.202 gml^-1 

= 83.19 mL

= 83.19 × 10^?3 L

Thus, molarity of the solution = (20/166) / 83.19 × 10^?3 L

= 1.45 M

c) Moles of KI

Moles of H₂O = 80/18 = 4.44

Thus, mole fraction of KI = moles of KI/ moles of KI + moles of water = 0.12/0.12 + 4.44

= 0.0263

2) Calculate the moles ratio as per the percentage

K 42.39 % therefore 42.39/ 39.09 =1.084

 Fe 15.21% therefore 56 15.21/ 56 = 0.27 

C 19.56% therefore 19.56 / 12 = 1.63

N 22.83% therefore 22.83 / 14 = 1.63

divide with the smallest

1.084/0.27 = 4

0.27/0/27 = 1

1.63/0.27 = 6

Therefore the formula is ​K​₄[Fe(CN)₆]