Question

Equal weights of $NaCl$ and $KCl$ are dissolved separately in equal volumes of solutions. Molarity of the two solutions will be :

• A. Equal
• B. That of NaCl will be less than that of KCl
• C. That of NaCl will be more than that of KCl
• D. That of NaCl will be about half of that of KCl solution

Answer 1

Answer: (C)

$M _{ KCl }=39+35.5=74.5$
Equal weight $74.5 g$
$n _{ NaCl }=\frac{74.5}{58.5}=1.27$
$n _{ NaCl }=\frac{74.5}{74.5}=1 mole$
Molarity $=\frac{\text { number of moles }}{ V }$
$=\frac{ n }{1 \text { liter }}$
$n _{ NaCl }> n _{ KCl }: M _{ NaCl }> M _{ KCl }$