We are given: - Molarity (M) = 3 M (mol/L) - Density of solution = 1.25 g/mL = 1250 g/L
The molality (m) is given by the formula: \[ m = \frac{\text{moles of solute}}{\text{kg of solvent}} \]
From molarity, the number of moles of NaCl in 1 liter of solution is: \[ \text{moles of NaCl} = 3 \, \text{mol} \]
To find the mass of NaCl, use the molar mass of NaCl (58.44 g/mol): \[ \text{mass of NaCl} = 3 \, \text{mol} \times 58.44 \, \text{g/mol} = 175.32 \, \text{g} \]
Now, the mass of the solvent (water) is: \[ \text{mass of solvent} = 1250 \, \text{g} - 175.32 \, \text{g} = 1074.68 \, \text{g} = 1.07468 \, \text{kg} \]
Thus, the molality is: \[ m = \frac{3 \, \text{mol}}{1.07468 \, \text{kg}} = 2.79 \, \text{m} \]
Therefore, the correct answer is 2.79 m.
A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
The equivalent resistance between the points \(A\) and \(B\) in the given circuit is \[ \frac{x}{5}\,\Omega. \] Find the value of \(x\). 
Method used for separation of mixture of products (B and C) obtained in the following reaction is: 