Volume of 3 M NaOH (formula weight 40 g m | vedclass

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(B) The molarity $M$ is defined as the number of moles of solute per liter of solution.$M = \frac{n_{NaOH}}{V_{sol} 	ext{ (in } L)}$First,calculate t

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(B) The molarity $M$ is defined as the number of moles of solute per liter of solution.$M = \frac{n_{NaOH}}{V_{sol} 	ext{ (in } L)}$First,calculate the number of moles of $NaOH$:$n_{NaOH} = \frac{	ext{mass}}{	ext{molar mass}} = \frac{84 \ g}{40 \ g \ mol^{-1}} = 2.1 \ mol$Now,use the molarity formula to find the volume $V$:$3 \ M = \frac{2.1 \ mol}{V 	ext{ (in } L)}$$V = \frac{2.1}{3} \ L = 0.7 \ L$Since $1 \ L = 1 \ dm^3$,$V = 0.7 \ dm^3 = 7 	imes 10^{-1} \ dm^3$.Thus,the value is $7$.

Volume of $3 \ M \ NaOH$ (formula weight $40 \ g \ mol^{-1}$) which can be prepared from $84 \ g$ of $NaOH$ is $ . . . . . . \times 10^{-1} \ dm^3$.

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