QJEEM24Ch15c5: JEE Main 2024 Chemistry - Class 11 JEEM24Ch15c5 Solution: Volume of (formula weight ) which can be prepared from of is…

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 $40\ \text{g/mol}$ .

The number of moles (n) is calculated using the formula:

So for our case:

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:

Where: C = concentration in molarity (M) n = number of moles V = volume in liters (L) - Note that $1\ \text{dm}^3 = 1\ \text{L}$ .

Since we want to find the volume (V), we can rearrange the formula to solve for V:

Using the moles of NaOH and the concentration for the preparation:

Calculate the volume:

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:

To express this volume as $\times 10^{-1}\ \text{dm}^3$ , we can write:

Therefore, the volume of $3\ \text{M}$ NaOH solution which can be prepared from $84\ \text{g}$ of NaOH is $7 \times 10^{-1}\ \text{dm}^3$ .