At 298 text{ K}, the molar conductivity of x% | vedclass

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(B) Step $1$: Calculate the conductivity $(\kappa)$ of the solution.$\kappa = G 	imes (l/A) = 1.9 	imes 10^{-3} 	ext{ S} 	imes 1.3 	ext{ cm}^{-1}

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(B) Step $1$: Calculate the conductivity $(\kappa)$ of the solution.$\kappa = G 	imes (l/A) = 1.9 	imes 10^{-3} 	ext{ S} 	imes 1.3 	ext{ cm}^{-1} = 2.47 	imes 10^{-3} 	ext{ S cm}^{-1}$.Step $2$: Calculate the molarity $(M)$ of the solution using the molar conductivity formula.$\Lambda_m = (\kappa 	imes 1000) / M$$123.5 = (2.47 	imes 10^{-3} 	imes 1000) / M$$M = 2.47 / 123.5 = 0.02 	ext{ mol L}^{-1}$.Step $3$: Calculate the mass of the solute in $1 	ext{ L}$ of solution.Since density is $1.0 	ext{ g mL}^{-1}$, $1 	ext{ L}$ of solution weighs $1000 	ext{ g}$.Mass of $MX = 	ext{moles} 	imes 	ext{molar mass} = 0.02 	ext{ mol} 	imes 75 	ext{ g mol}^{-1} = 1.5 	ext{ g}$.Step $4$: Calculate the percentage by weight $(w/w)$.$\% (w/w) = (	ext{mass of solute} / 	ext{mass of solution}) 	imes 100 = (1.5 / 1000) 	imes 100 = 0.15$.$0.15 = 15 	imes 10^{-2}$.Therefore, the value of $x$ is $15$.

Ion and $\lambda ^o / (S \ cm^2 \ mol^{-1})$	Ion and $\lambda ^o / (S \ cm^2 \ mol^{-1})$
$H^{+} : 349.6$	$OH^{-} : 199.1$
$Na^{+} : 50.1$	$Cl^{-} : 76.3$
$K^{+} : 73.5$	$Br^{-} : 78.1$
$Ca^{2+} : 119.0$	$CH_3COO^{-} : 40.9$
$Mg^{2+} : 106.0$	$SO_4^{2-} : 160.0$

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