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.
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)
Help me to solve this problem. 4.0 gm of NaOH are contained in one decilitre of solution. Its molarity would be: (a) 4 M (b)2 M (c)1 M (d)1.5 M
why o-toluidine is a weaker base than aniline?
Carbon compounds containing more than one halogen atom are called polyhalogen compounds. Most of these compounds are valuable in industry and agriculture. Some important polyhalogen compounds are described as follows: Q : The Liver Is Responsible For Much Of The Liver Is Responsible For Much Of The Pentose Phosphate Activity Explain
The Liver Is Responsible For Much Of The Pentose Phosphate Activity Explain
Vapour pressure of methanol in water Give me answer of this question. An aqueous solution of methanol in water has vapour pressure: (a) Equal to that of water (b) Equal to that of methanol (c) More than that of water (d) Less than that of water
The accuracy of your written English will be taken into account in marking. 1. (a) Identify the spectator ions in the following equation &nb
Illustrate is HCl an acid or a base ?
At 20°C the solubility of solid naphthalene in hexane is 0.09 mol/mol of solution. Use this information and the data below to estimate the following for this system: a) The mol fraction of naphthalene in the vapour phase in equ
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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