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

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that of NaCl will be more than that of KCI solution

that of NaCl will be half than that of KCl solution

equal

that of NaCl will be less than that of KCI solution

answer is C.

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Detailed Solution

$M_{kcl} = 39 + 35.5 = 74.5$

Equal weight $74.58$

$n_{Ncl} = \frac{74.5}{58.5} = 127$

$m_{nacl} = \frac{74 .5}{74 .5} = 1 mole$

$Molarity = \frac{number of moles}{V}$

$= \frac{n}{1 lite r}$

$n_{NaCl} > n_{KCl} : M_{NaCl} > M_{KCl}$

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