Molarity of cane sugar solution
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.
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.
What is the reason behind this that some medications contain hcl?
The total angular momentum of an atom includes an electron spin component as well as an orbital component.The orbital motion of each electron of an atom contributes to the angular momentum of the atom, as described earlier. An additional
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
Choose the right answer from following. The molality of 90% H2SO4 solution is: [density=1.8 gm/ml] (a)1.8 (b) 48.4 (c) 9.18 (d) 94.6
Illustrate HCl is polar or non-polar?
Which one of the following pairs of solutions can we expect to be isotonic at the same temperature:(i) 0.1M Urea and 0.1M Nacl (ii) 0.1M Urea and 0.2M Mgcl2 (iii) 0.1M Nacl and 0.1M Na2SO4 (iv) 0.1M Ca(NO3<
Benzene and toluene form nearly ideal solutions. At 20°C, the vapour pressure of benzene is 75 torr and that of toluene is 22 torr. The parial vapour pressure of benzene at 20°C for a solution containing 78g of benzene and 46g of toluene in torr is: (a) 50 (b)
Illustrate what is protein in Chemistry?
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
If 20ml of 0.4N, NaoH solution completely neutralises 40ml of a dibasic acid. The molarity of the acid solution is: (a) 0.1M (b) 0.2M (c) 0.3M (d) 0.4M Choose the right answer fron above.
18,76,764
1924025 Asked
3,689
Active Tutors
1424740
Questions Answered
Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!