Sugar solution
The solution of sugar in water comprises: (i) Free atoms (ii) Free ions (iii) Free molecules (iv) Free atom and molecules. Choose the right answer from the above.
The solution of sugar in water comprises: (i) Free atoms (ii) Free ions (iii) Free molecules (iv) Free atom and molecules.
Choose the right answer from the above.
These types of polymers are formed as a result of condensation reaction between monomer units. Some common examples are being discussed here: 1. Polyesters Q : Molar mass of solute The boiling point The boiling point of benzene is 353.23 K. If 1.80 gm of a non-volatile solute was dissolved in 90 gm of benzene, the boiling point is increased to 354.11 K. Then the molar mass of the solute is: (a) 5.8g mol-1 (b)
The boiling point of benzene is 353.23 K. If 1.80 gm of a non-volatile solute was dissolved in 90 gm of benzene, the boiling point is increased to 354.11 K. Then the molar mass of the solute is: (a) 5.8g mol-1 (b)
1. Physical state: the first three aliphatic acids are colourless liquids with pungent smell. The next six are oily liquids with an odour of rancid butter while the higher members are colourless, odourless waxy solids. Benzoic acid is referred to
Help me to go through this problem. Molecular weight of glucose is 180. A solution of glucose which contains 18 gms per litre is : (a) 2 molal (b) 1 molal (c) 0.1 molal (d)18 molal
why pentahalids are more covalent than tetrahalids
What will be the molality of a solution containing 18g of glucose (having mol. wt. = 180) dissolved in 500g of water: (i) 1m (ii) 0.5m (iii) 0.2m (iv) 2m
Three dimensional applications of the Schrodinger equation are introduced by the particle-in-a-box problem.So far only a one-dimensional problem has been solved by application of the Schrodinger equation. Now the allowed energies and the probability functi
The molality of a solution will be: (i) Number of moles of solute per 1000 ml of solvent (ii) Number of moles of solute per 1000 gm of solvent (iii) Number of moles of solute per 1000 ml of solution (iv) Number of gram equivalents of solute per 1000 m
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
Nitrogen tetroxide (melting point: -11.2°C, normal boiling point 21.15°C) decomposes into nitrogen dioxide according to the following reaction: N2O4(g) ↔ 2 NO2(g)<
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