Problem on mole fraction of glucose
Provide solution of this question. While 1.80gm glucose dissolve in 90 of H2O , the mole fraction of glucose is: (a) 0.00399 (b) 0.00199 (c) 0.0199 (d) 0.998
The accuracy of your written English will be taken into account in marking. 1. (a) Identify the spectator ions in the following equation &nb
The equilibrium constant can be treated as a particular type of molecular distribution. Consider the simplest gas-phase reaction, one in which molecules of A are converted to molecules of B. the reaction, described by the equation Q : Problem on moles of solution The number The number of moles of a solute in its solution is 20 and total no. of moles are 80. The mole fraction of solute wil be: (a) 2.5 (b) 0.25 (c) 1 (d) 0.75
The number of moles of a solute in its solution is 20 and total no. of moles are 80. The mole fraction of solute wil be: (a) 2.5 (b) 0.25 (c) 1 (d) 0.75
Help me to go through this problem. Molarity of a solution containing 1g NaOH in 250ml of solution: (a) 0.1M (b) 1M (c) 0.01M (d) 0.001M
Help me to go through this problem. Lowering of vapour pressure is highest for: (a) urea (b) 0.1 M glucose (c) 0.1M MgSo4 (d) 0.1M BaCl2
In these reaction oxygen atom of carbonyl group is replaced by either one divalent group or two monovalent groups. Reaction by ammonia derivatives: aldehydes and ketones react with a number of ammonia derivatives such as hydroxylaminem hydrazine, semicarbazide etc. in weak acidic medium.
20 g of hydrogen is present in 5 litre of vessel. Determine he molar concentration of hydrogen: (a) 4 (b) 1 (c) 3 (d) 2 Choose the right answer from above.
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<
The distribution law is exerted for the distribution of basic acid among: (i) Water and ethyl alcohol (ii) Water and amyl alcohol (iii) Water and sulphuric acid (iv) Water and liquor ammonia What is the right answer.
The first method begins with a well defined layer, or boundary, of solution near the center of rotation and tracks the movement of this layer to the outside of the cell as a function of time. Such a method is termed a sedimentary velocity experiment. A
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