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You weight 14,5 gramm KOH in a bottle and than fill that bottle with deionized water till it holds 500ml.
Dissolve 1.0 mole gas (17 gram) in 1.0 Liter water
40 grams, this is the 1M NaOH standard laboratory solution.
Sulphuric Acid requires, I believe 27.2ml to make a 1N solution.
to prepare 1N we have to dilute 40gms of NaOH in 1 litre of water as for NaOH normality =molarity so to prepare 0.1N NaOH we have to dilute 4gms of NaOH in 1 litre of water..
You weight 14,5 gramm KOH in a bottle and than fill that bottle with deionized water till it holds 500ml.
Dissolve 1.0 mole gas (17 gram) in 1.0 Liter water
Dissolve 1.0 mole gas (17 gram) in 1.0 Liter water
40 grams, this is the 1M NaOH standard laboratory solution.
Sulphuric Acid requires, I believe 27.2ml to make a 1N solution.
to prepare 1N we have to dilute 40gms of NaOH in 1 litre of water as for NaOH normality =molarity so to prepare 0.1N NaOH we have to dilute 4gms of NaOH in 1 litre of water..
weigh out 365g of HCl pellets and dilute to 1 liter to prepare the 10N HCl solution. Reasons:N is short for NORMAL SOLUTIONS, The definition of a NORMAL SOLUTION is a solution that contains 1 gram equivalent weight (gEW) per liter solution. An equivalent weight is equal to the molecular weight divided by the valence (replaceable H ions). eg:1N NaCl = 58.5 g/L 1N HCl = 36.5 g/L 1N H2S04 = 49 g/L Problems involving normality are worked the same as those involving molarity but the valence must be considered: 1N HCL the MW= 36.5 the EW = 36.5 and 1N would be 36.5g/L 1N H2SO4 the MW = 98 the EW = 49 and 1N would be 49 g/L 1N H3PO4 the MW = 98 the EW = 32.7 and 1N would be 32.7 g/L so,u can weigh out 365g of HCl pellets and dilute to 1 liter to prepare the 10N HCl solution.
It is possible only if you evaporate the water.
Dissolve 16.235 g of ICl into 1000 ml solution with glacial acetic acid. The solution will be 0.1N Wij's Solution. However weighing is a very difficult task you need to have the required skill and sound analytical knowledge and detailed MSDS of ICl. Man Katuwal
Take specific volume of 3N solution and increase the volume three times by adding distilled water.
1/1n
Preparing 1N HCl for 1L. 1N=1M in HCl. Conc. HCl= 12M M1V1=M2V2 12*V1=1*1000 V1=1000/12 V1=83.33ml 1N HCl= 83.33ml of Conc. HCl in 1L of water 2N HCl= 167ml of Conc. HCl in 1L of water.