Colligative property associated question
Give me answer of this question. Which of the following is not a colligative property : (a)Optical activity (b)Elevation in boiling point (c)Osmotic pressure (d)Lowering of vapour pressure
Can someone please help me in getting through this problem. 2.0 molar solution is acquired, when 0.5 mole solute is dissolved in: (i) 250 ml solvent (ii) 250 g solvent (iii) 250 ml solution (iv) 1000 ml solvent
Among the below shown which crystal will be soft and have low melting point: (a) Covalent (b) Ionic (c) Metallic (d) MolecularAnswer: (d) Molecular crystals are soft and have low melting point.
Can someone please help me in getting through this problem. Determine the normality of a solution having 4.9 gm H3PO4 dissolved in 500 ml water: (a) 0.3 (b) 1.0 (c) 3.0 (d) 0.1
Freezing point of a substance is the temperature at which solid and liquid phases of the substance coexist. It is defined as the temperature at which its solid and liquid phases have the same vapour pressure. The freezing point o
Describe how dipole attractions, London dispersion forces and the hydrogen bonding identical?
Select the right answer of the question. If 18 gm of glucose (C6H12O6) is present in 1000 gm of an aqueous solution of glucose, it is said to be: (a)1 molal (b)1.1 molal (c)0.5 molal (d)0.1 molal
Catalyst is a substance which accelerates the rate of a chemical reaction without undergoing any change in its chemical composition or mass during the reaction. The phenomenon of increasing the rate of a reaction with the help of a catalyst is known as catalysis.
Select the right answer of the question. Which of the following is not a colligative property : (a) Osmotic pressure (b) Elevation in B.P (c) Vapour pressure (d) Depression in freezing point
Give me answer of this question. What weight of ferrous ammonium sulphate is requiored to prepare 100 ml of 0.1 normal solution (mol. wt. 392): (a) 39.2 gm (b) 3.92 gm (c)1.96 gm (d)19.6 gm
The number of molecular orbitals and molecular motions of each symmetry type can be deduced. Let us continue to use the C2v point group and the H2O molecule to illustrate how the procedure develop
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