Question

Calculate the molarity molality and mole fraction of ki if density is 1.202 by weight 20%

Answer 1

Mass % is given so that take total mass = 100 g

Mass of KI = 20 % of 100 = 20 g

Mass of water = total – mass of KI = 100 – 20 = 80 g of water

Mass of solvent (water) in kg = 80 /1000 = 0.08 Kg

Molar mass of KI = 39 + 127 = 166 g mol−1

Use the formula

Number of moles of KI = 20 / 166 = 0.1205 mol

Use the formula

Molality of KI = 0.1205 / 0.08 = 1.51 m

(b)

Given that the density of the solution = 1.202 g mL−1

Volume of 100 g solution = 100 / 1.202 = 83.2 mL

Divide by 1000 to convert in liter we get = 83.2/1000 =0.0832 liter

Use the formula

Answer 2

$\huge\underline{\bold{\green{Answer :-}}}$

a. Molality = 1.5 m.

b. Molarity = 1.44 M.

c. Mole fraction = 0.0263.

★$\underline{\bold{\pink{Given :}}}$

20% solution of KI means 20 g of KI are present in 100 g of solution or 80 g of water.

Density of KI in aqueous solution = $\bold{1.202\:g\:mL^{-1}}$

Mass of KI = $\bold{20 g}$

★$\underline{\bold{\orange{To\: find :}}}$

a. Molality

b. Molarity

c. Mole fraction

★$\underline{\bold{\red{How\: to \:Find :}}}$

$\rm{Molar\: mass\: of\: <strong>KI</strong><strong> </strong>\:}= \bold{39 + 127 = 166\:g\:mol^{-1}}$

$\rm{Moles\: of\: <strong>KI</strong>} = \bold{\frac{20}{166} = 0.120}$

$\bold{\rm{Therefore,}}$

$\rm{A.\:<strong> </strong>Molality =} \bold{\frac{0.120}{80}\times{1000} = \huge 1.5\:m}$

$\rm{<strong> </strong>B.<strong> </strong>\:Volume\: of \:solution =} \bold{\frac{100}{1.202} = 83.19\:mL}$

$\therefore \: \rm{ Molarity,}$

$\implies \bold{\frac{0.120}{83.19}\times{1000} = \huge 1.44\:M}$

$\rm{C. \:Moles\:of\:KI = }\bold{0.120}$

$\rm{Moles\:of\:H_2O = } \bold{\frac{80}{18} = 4.44}$

$\therefore \: \rm{Mole\:fraction, }$

$\implies \bold{x_{KI} = \frac{0.120}{4.44 + 0.120} = \huge 0.0263}$