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

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

Tardigrade Admin · Accepted Answer

M KCl =39+35.5=74.5 Equal weight 74.5 g n NaCl =(74.5/58.5)=1.27 n NaCl =(74.5/74.5)=1 mole Molarity =( text number of moles / V ) =( n /1 text liter ) n NaCl > n KCl: M NaCl > M KCl

Q. Equal weights of $N a Cl$ and $K Cl$ are dissolved separately in equal volumes of solutions. Molarity of the two solutions will be :

Equal

That of NaCl will be less than that of KCl

That of NaCl will be more than that of KCl

That of NaCl will be about half of that of KCl solution

$M_{K Cl} = 39 + 35.5 = 74.5$
Equal weight $74.5 g$
$n_{N a Cl} = \frac{74.5}{58.5} = 1.27$
$n_{N a Cl} = \frac{74.5}{74.5} = 1 m o l e$
Molarity $= \frac{number of moles}{V}$
$= \frac{n}{1 liter}$
$n_{N a Cl} > n_{K Cl} : M_{N a Cl} > M_{K Cl}$

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