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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Solutions
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Match the column-I with column-II ,which are true for 20% w/w aqueous KI solution (density of KI solution is 1.20g/ml & M.wt of KI is 166 g/mol)

Column-I Column-II
(1) Molarity of KI solution (p) 1.2mol$\; Kgmol^{-1}$
(2) Normality of KI solution (q) 1.4eq$\; L^{-1}$
(3) Molality of KI solution (r) 0.026
(4) Mole fraction of KI (s) 1.4mol$L^{-1}$

 

$\begin{array}{1 1}(a)\;(1)\rightarrow (p),(2)\rightarrow (r),(3)\rightarrow (q),(4)\rightarrow (s)\\(b)\;(1)\rightarrow (s),(2)\rightarrow (q),(3)\rightarrow (r),(4)\rightarrow (p)\\(c)\;(1)\rightarrow (s),(2)\rightarrow (q),(3)\rightarrow (p),(4)\rightarrow (r)\\(d)\;(1)\rightarrow (q),(2)\rightarrow (s),(3)\rightarrow (p),(4)\rightarrow (r)\end{array}$

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1 Answer

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Toolbox:
  • $Molality =\large \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}$
  • $Molarity =\large \frac{\text{Moles of solute}}{\text{Volume of solution in Litres}}$
  • The normality of a solution is the gram equivalent weight of a solute per liter of solution.
Given, molecular weight of KI is 166 g/mol
and, density of KI solution is 1.20 g/mL
Molarity of the KI solution =$\large\frac{weight}{molecular\; weight}\times\frac{1000}{V(mL)}$$\times density \; of\; KI \; solution$
$= \large\frac{1.20\times \Large\frac{20}{100}}{166\times 1}$$\times 1000 = 1.4 \;M$
Since, the normality of a solution is the gram equivalent weight of a solute per liter of solution. Thus, Normality of the KI solution = Molarity of the KI solution
$Molality =\large \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}$
Molality $=\large\frac{20}{166\times 0.1}$
$\Rightarrow 1.2$
Hence (c) is the correct answer.
answered Feb 14, 2014 by sreemathi.v
edited Jul 20, 2014 by mosymeow_1
 
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