Problem based on lowering in vapour pressure
Help me to solve this problem. An aqueous solution of glucose was prepared by dissolving 18 g of glucose in 90 g of water. The relative lowering in vapour pressure is: (a) 0.02 (b)1 (c) 20 (d)180
i) Show that the equilibrium constant Kp for the reaction CaCo3(s) ↔ CaO(s) +CO2(g)is about unity (i.e. = 1.0) at 895 °C.ii) If two grams of calcium carbonate are pl
what are the most important inorganic molecular substances for living beings?
Wax is an example of: (a) Ionic crystal (b) Covalent crystal (c) Metallic crystal (d) Molecular crystalAnswer: (d) Iodine crystals are molecular crystals, in which constituent particles are molecules having inter particle
Ionic dissociation depends on the dielectric constant of the solvent.The Arrhenius that ions are in aqueous solutions in equilibrium with parent molecular species allows many of the properties of ionic solutions to be understood. But difficulties began to
living beings are made up of organic and inorganic substances.according to their complexity of their molecules how can ach of these substances be classified?
171 g of cane sugar (C12H22O11) is dissolved in one litre of water. Find the molarity of the solution: (i) 2.0 M (ii) 1.0 M (iii) 0.5 M (iv) 0.25 M Choose the right answer from above.
Chromium(III) hydroxide is highly insoluble in distilled water but dissolves readily in either acidic or basic solution. Briefly explain why the compound can dissolve in acidic or in basic but not in neutral solution. Write appropriate equations to support your answer.
The temperature reliance of internal energy and enthalpy depends on the heat capacities at constant volume and constant pressure. The internal energy and enthalpy of chemical systems and the energy changes that accompany chemical reactions depend on the
Karl Scheele, the Swedish chemist, was
When 5.85 g of NaCl (having molecular weight 58.5) is dissolved in water and the solution is prepared to 0.5 litres, the molarity of the solution is: (i) 0.2 (ii) 0.4 (iii) 1.0 (iv) 0.1
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