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.
According to the generally accepted definition of the ideal solution there are equal interaction forces acting between molecules belonging to the same or different species. (This is equivalent to the statement that the activity of the components equals the concentration.) Strictly speaking, this concept is valid in ecological systems (isotopic mixtures of an element, hydrocarbons mixtures, etc.). It is still usual to talk about ideal solutions as limiting cases in reality since very dilute solutions behave ideally with respect to the solvent. This law is further supported by the fact that Raoult’s law empirically found for describing the behaviour of the solvent in dilute solutions can be deduced thermodynamically via the assumption of ideal behaviour of the solvent.
Answer the following questions:
(a) Give one example of miscible liquid pair which shows negative deviation from Raoult’s law. What is the reason for such deviation?
(b) (i) State Raoult’s law for a solution containing volatile components.
OR
(ii) Raoult’s law is a special case of Henry’s law. Comment.
(c) Write two characteristics of an ideal solution.
If \[ f(x) = \int \frac{1}{x^{1/4} (1 + x^{1/4})} \, dx, \quad f(0) = -6 \], then f(1) is equal to: