The density of 3 M solution of NaCl is 1.0 | vedclass

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Question

(A) The molarity $(M)$ is $3 \ M$ and density $(d)$ is $1.0 \ g \ mL^{-1}$.The molar mass of $NaCl$ is $23 + 35.5 = 58.5 \ g \ mol^{-1}$.The formula f

Vedclass · Accepted Answer

$364$

Vedclass · Answer

(A) The molarity $(M)$ is $3 \ M$ and density $(d)$ is $1.0 \ g \ mL^{-1}$.The molar mass of $NaCl$ is $23 + 35.5 = 58.5 \ g \ mol^{-1}$.The formula for molality $(m)$ is $m = \frac{1000 	imes M}{1000 	imes d - M 	imes M_{solute}}$.Substituting the values: $m = \frac{1000 	imes 3}{1000 	imes 1 - 3 	imes 58.5} = \frac{3000}{1000 - 175.5} = \frac{3000}{824.5} \approx 3.6385 \ m$.Expressing in $10^{-2} \ m$: $3.6385 	imes 10^2 	imes 10^{-2} \approx 364 	imes 10^{-2} \ m$.The nearest integer is $364$.

The density of $3 \ M$ solution of $NaCl$ is $1.0 \ g \ mL^{-1}$. Molality of the solution is $...... \times 10^{-2} \ m$. (Nearest integer). Given: Molar mass of $Na$ and $Cl$ is $23$ and $35.5 \ g \ mol^{-1}$ respectively.

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