JEE Main 2024 — Some Basic Concepts Of Chemistry Question with Solution

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:

$$n=\frac{mass}{molarmass}$$

So for our case:

$$n=\frac{84\text{g}}{40\text{g/mol}}=2.1\text{mol}$$

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:

$$C=\frac{n}{V}$$

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:

$$V=\frac{n}{C}$$

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

$$V=\frac{2.1\text{mol}}{3\text{M}}$$

Calculate the volume:

$$V=\frac{2.1}{3}=0.7\text{L}$$

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:

$$V=0.7{\text{dm}}^3$$

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

$$V=7\times 10^{-1}{\text{dm}}^3$$

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