The density of '$x$' $\mathrm{M}$ solution ('$x$' molar) of $\mathrm{NaOH}$ is $1.12 \mathrm{~g} \mathrm{~mL}^{-1}$, while in molality, the concentration of the solution is $3 \mathrm{~m}$ ( 3 molal). Then $x$ is

(Given : Molar mass of $\mathrm{NaOH}$ is $40 \mathrm{~g} / \mathrm{mol}$)

<p>To find the molarity, $x$, of the $\mathrm{NaOH}$ solution, we need to relate the given density and molality. Molarity is defined as the number of moles of solute per liter of solution, while molality is defined as the number of moles of solute per kilogram of solvent. We are given the following:</p> <ul> <li>Molality (m) = $3\ \mathrm{mol/kg}$ (3 molal)</li> <li>Density of solution ($\rho$) = $1.12\ \mathrm{g/mL}$ = $1120\ \mathrm{g/L}$</li> <li>Molar mass of $\mathrm{NaOH}$ ($M_\mathrm{NaOH}$) = $40\ \mathrm{g/mol}$</li> </ul> <p>First, let's find the number of moles of $\mathrm{NaOH}$ in 1 kg of water, which is the definition of molality. Since the molality is 3 molal, we have:</p> <p>$3\ \mathrm{mol}$ of $\mathrm{NaOH}$ in 1 kg of water (or 1000 g of water).</p> <p>Now, to find the molarity, we need the volume of the solution in which these moles of $\mathrm{NaOH}$ are dissolved. The mass of 1 L (1000 mL) of solution, given the density, is:</p> <p>$1120\ \mathrm{g}$</p> <p>The mass of $\mathrm{NaOH}$ in 3 moles is:</p> <p>$3 \mathrm{mol} \times 40\ \mathrm{g/mol} = 120\ \mathrm{g}$</p> <p>Thus, the mass of water in this 1 L solution can be found by subtracting the mass of $\mathrm{NaOH}$ from the total mass of the solution:</p> <p>$1120\ \mathrm{g} - 120\ \mathrm{g} = 1000\ \mathrm{g}$</p> <p>So, we have 3 moles of $\mathrm{NaOH}$ in 1 L of solution. Therefore, the molarity (x) of the solution is directly 3 M (since molarity is moles of solute per liter of solution). Hence, $x = 3$.</p> <p>Therefore, the correct answer is Option B: 3.0.</p>