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

<p>First, let's calculate the number of moles of NaOH that can be prepared from $84\ \text{g}$ of NaOH. The molar mass of NaOH is given as $40\ \text{g/mol}$.</p> <p>The number of moles (n) is calculated using the formula:</p> $n = \frac{mass}{molar\ mass}$ <p>So for our case:</p> $n = \frac{84\ \text{g}}{40\ \text{g/mol}} = 2.1\ \text{mol}$ <p>Now that we know the number of moles, we can find out the volume of a $3\ \text{M}$ NaOH solution that can be prepared from it. The concentration (C) of a solution is related to the number of moles (n) and volume (V) by the following formula:</p> $C = \frac{n}{V}$ <p>Where: C = concentration in molarity (M) n = number of moles V = volume in liters (L) - Note that $1\ \text{dm}^3 = 1\ \text{L}$.</p> <p>Since we want to find the volume (V), we can rearrange the formula to solve for V:</p> $V = \frac{n}{C}$ <p>Using the moles of NaOH and the concentration for the preparation:</p> $V = \frac{2.1\ \text{mol}}{3\ \text{M}}$ <p>Calculate the volume:</p> $V = \frac{2.1}{3} = 0.7\ \text{L}$ <p>To convert this volume to $\text{dm}^3$ (which is equivalent to liters), we use the conversion factor $1\ \text{L} = 1\ \text{dm}^3$. Therefore:</p> $V = 0.7\ \text{dm}^3$ <p>To express this volume as $\times 10^{-1}\ \text{dm}^3$, we can write:</p> $V = 7 \times 10^{-1}\ \text{dm}^3$ <p>Therefore, the volume of $3\ \text{M}$ NaOH solution which can be prepared from $84\ \text{g}$ of NaOH is $7 \times 10^{-1}\ \text{dm}^3$. </p>