NEET Practice Questions (MCQs) with Answers & Solutions

The solubility of COâ‚‚ in water decreases with an increase in temperature. This is because the dissolution of COâ‚‚ in water is an exothermic process, and according to Le Chatelier's principle, an increase in temperature shifts the equilibrium towards the endothermic direction, decreasing the solubility.

For an ideal solution, the mole fraction of a component in the vapor phase is directly proportional to its vapor pressure. Since the vapor pressures of A and B are 125.0 and 62.5 mm, respectively, their mole fractions in the vapor phase will be in the ratio of 2:1, which corresponds to 0.75 and 0.25.

$ p_A = p^0 _A \times X_A $ $ = 125 \times 0.6 $ = 75 mm $ p_B = p ^ 0_B \times X_B = 62.5 \times 0.4 = 25mm $ $ p_{total} = p_A + p_B = 75 + 25 = 100 mm $ $ p_A = p_{Total} \times Y_A $ $ \therefore 75 = 100 \times Y_A $ $ \therefore Y_A = 0.75 $ $ Similarly \therefore Y_B = 0.25 $

Ideal solutions are formed by mixing two liquids that are chemically similar and have similar intermolecular forces. Benzene and toluene are both non-polar, aromatic hydrocarbons, and their molecules have similar sizes and shapes, allowing them to mix easily, forming an ideal solution.

$ For NaCl value of i= 2.0 $ $ \therefore \triangle Tb = i.m.Kb = 2 \times 0.1 \times Kb $ $ \therefore \triangle Tb = { Kb \over 5} $

Cellophane is a semi-permeable membrane made from regenerated cellulose, which allows the passage of small molecules like water but prevents the passage of larger molecules, making it an effective semi-permeable membrane for various applications.

$ X_A = 1 -X_B = 1 - 0.4 = 0.6 $ $ Y_A = 1 - Y_B = 1- 0.25 = 0.75 $ $ p_B = p^ \ circ _B \times X_B $ $ \therefore p_B = 40 \times 0.4 $ = 16 mm $ p_B = P_{total} \times Y_B $ $ \therefore 16 = P_{total} \times 0.25 $ $ \therefore P_{total} = 64 mm $ $ p_A = P_{total} \times Y_A = 64 \times 0.75 = 48 mm $ $ Now p_A = p ^ \circ _A \times X_A $ $ 48 = p ^ \circ _A \times 0.6 $ $ \therefre p ^ \circ _A =80 mm $

Liquid present in RBC is isotonic with 0.91 % w/v solution of NaCl $ \therefore Morality of soluble particles in 0.91 \% w/v NaCl Solution $ $ = { 2 \times 1000 \times 0.91 \over 58.5 \times 100 } $ $ = 2 \times 0.1555 $ = 0.311 M

To determine whether a solution is hypotonic or hypertonic compared to the fluids inside red blood cells (RBCs), we need to consider the concentration of solute particles. The normal osmotic pressure inside RBCs corresponds to a 0.9% NaCl solution, which is approximately 0.154 M NaCl. Therefore, any solution with a lower concentration than 0.154 M NaCl will be hypotonic to RBCs.