The density of 'x' M solution ('x' molar) of | vedclass

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(B) The relationship between molality $(m)$ and molarity $(M)$ is given by the formula:$m = \frac{1000 	imes M}{(1000 	imes d) - (M 	imes Mw_{	ext

Vedclass · Accepted Answer

$3.0$

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(B) The relationship between molality $(m)$ and molarity $(M)$ is given by the formula:$m = \frac{1000 	imes M}{(1000 	imes d) - (M 	imes Mw_{	ext{solute}})}$Given:$m = 3 \ molal$$d = 1.12 \ g \ mL^{-1}$$Mw_{	ext{solute}} = 40 \ g \ mol^{-1}$$M = x$Substituting the values into the formula:$3 = \frac{1000 	imes x}{(1000 	imes 1.12) - (x 	imes 40)}$$3 = \frac{1000x}{1120 - 40x}$$3(1120 - 40x) = 1000x$$3360 - 120x = 1000x$$3360 = 1120x$$x = \frac{3360}{1120} = 3$Therefore,the value of '$x$' is $3$.

The density of '$x$' $M$ solution ('$x$' molar) of $NaOH$ is $1.12 \ g \ mL^{-1}$. While in molality,the concentration of the solution is $3 \ m$ $(3 \ molal)$. Then '$x$' is.
(Given: Molar mass of $NaOH$ is $40 \ g \ mol^{-1}$)

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The density of '$x$' $M$ solution ('$x$' molar) of $NaOH$ is $1.12 \ g \ mL^{-1}$. While in molality,the concentration of the solution is $3 \ m$ $(3 \ molal)$. Then '$x$' is.(Given: Molar mass of $NaOH$ is $40 \ g \ mol^{-1}$)