What is the molarity of a solution prepared by dissolving 5.85 g of NaCl in 500 mL of water?
(Molar mass of NaCl = 58.5 g/mol)
- Calculate the number of moles of NaCl: \[ \text{moles} = \frac{5.85\, \text{g}}{58.5\, \text{g/mol}} = 0.1\, \text{mol} \] - Volume of solution = 500 mL = 0.5 L - Molarity (M) = \(\frac{\text{moles}}{\text{volume in liters}} = \frac{0.1}{0.5} = 0.2\, \text{M}\)
Answer: \(\boxed{\text{B}}\)
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.