$\text{Molality of 0.8 M H}_2\text{SO}_4 \text{ solution (density 1.06 g cm}^{-3}\text{) is \_\_\_\_\_\_} \times 10^{-3} \, \text{m}.$
Correct Answer: 815
Solution and Explanation
To calculate the molality, we need the mass of the solvent in kilograms and the moles of \( \text{H}_2\text{SO}_4 \).
- Given molarity (\( M \)) of \( \text{H}_2\text{SO}_4 \): \( 0.8 \, \text{mol/L} \).
- Density of solution = \( 1.06 \, \text{g/cm}^3 \).
- Molar mass of \( \text{H}_2\text{SO}_4 \) = \( 98 \, \text{g/mol} \).
Step 1. Calculate the mass of 1 L of solution:
\(\text{Mass of solution} = 1.06 \times 1000 = 1060 \, \text{g}\)
Step 2. Calculate the moles of \( \text{H}_2\text{SO}_4 \) in 1 L of solution:
\(\text{Moles of } \text{H}_2\text{SO}_4 = 0.8 \, \text{mol}\)
Step 3. Calculate the mass of \( \text{H}_2\text{SO}_4 \):
\(\text{Mass of } \text{H}_2\text{SO}_4 = 0.8 \times 98 = 78.4 \, \text{g}\)
Step 4. Calculate the mass of the solvent (water) in the solution:
\(\text{Mass of water} = 1060 - 78.4 = 981.6 \, \text{g} = 0.9816 \, \text{kg}\)
Step 5. Calculate the molality (\( m \)):
\(m = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} = \frac{0.8}{0.9816} \approx 0.815 \, \text{mol/kg}\)
Step 6. Convert to \( \times 10^{-3} \) scale:
\(\text{Molality} = 815 \times 10^{-3} \, m\)
The Correct Answer is: 815
Top Questions on Solutions
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.- A solution of glucose (molar mass = 180 g mol\(^{-1}\)) in water has a boiling point of 100.20°C. Calculate the freezing point of the same solution. Molal constants for water \(K_f\) and \(K_b\) are 1.86 K kg mol\(^{-1}\) and 0.512 K kg mol\(^{-1}\) respectively.
- Define Azeotrope. What type of Azeotrope is formed by negative deviation from Raoult's law? Give an example.
- Give reasons:
(a) Cooking is faster in a pressure cooker than in an open pan.
(b) On mixing liquid X and liquid Y, volume of the resulting solution decreases. What type of deviation from Raoult's law is shown by the resulting solution? What change in temperature would you observe after mixing liquids X and Y? - An unripe mango placed in a concentrated salt solution to prepare pickle, shrivels because __________
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