Friendly Nerd - Exercise of Chemistry 11
Dissolved ammonium ammonium sulfate, (NH 4) 2 SO 4 in water, ammonium having obtained ammonium a solution whose concentration of ammonium ion was 2.00 mol / dm 3. Given that the density of the solution was 1.12 g / cm 3 to calculate the molar fraction of the solute in the solution. Additional Information:
Well, the measures are presented in different scales, which leads us to equate them: 2mol/dm -> 2mol / L (1DM = 1L) 1.12 g / cm -> 1.120g / L (1cm = 1ml, and move to G, just multiply by 10 ) Now the question: What is the mass of solute mole? Precissamos to the equation that separates the first ion (NH4) 2SO4-> 2NH4 + SO4 | (NH4) 2SO4 = 132g/mol and 36g/mol 2NH4 = | ie the solute was exactly 1 mol for its mass to form 2mol / L NH4 out of 132g / L = 1 mol Now another question: What is the mass of the solution ammonium in moles? We have that the volumetric mass of the solution is 1.120g / L. Except however, to that extent we have the mass of H20 and (NH4) 2SO4. SO, just how much longer we draw solute ammonium and ascertain how much leftover dough as this will be the mass of the water and how well we know its sum of MOLS. 1120g - 132g = 988G H2O | H2O = 18g/mol | 988/18 ~ 55 moles H20 FINALLY just calculate the standard formula substituting the values: x = x1/x1 + x2 WE HAVE X = 1 mol + 1 mol SO 55mol X ~ 0.02 * NOTE: it follows the rule that 0 <x <1, also taking the sum of Molara fraction of the solute + water = 1 correctly following the rule concerned Reply
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Dissolved ammonium ammonium sulfate, (NH 4) 2 SO 4 in water, ammonium having obtained ammonium a solution whose concentration of ammonium ion was 2.00 mol / dm 3. Given that the density of the solution was 1.12 g / cm 3 to calculate the molar fraction of the solute in the solution. Additional Information:
Well, the measures are presented in different scales, which leads us to equate them: 2mol/dm -> 2mol / L (1DM = 1L) 1.12 g / cm -> 1.120g / L (1cm = 1ml, and move to G, just multiply by 10 ) Now the question: What is the mass of solute mole? Precissamos to the equation that separates the first ion (NH4) 2SO4-> 2NH4 + SO4 | (NH4) 2SO4 = 132g/mol and 36g/mol 2NH4 = | ie the solute was exactly 1 mol for its mass to form 2mol / L NH4 out of 132g / L = 1 mol Now another question: What is the mass of the solution ammonium in moles? We have that the volumetric mass of the solution is 1.120g / L. Except however, to that extent we have the mass of H20 and (NH4) 2SO4. SO, just how much longer we draw solute ammonium and ascertain how much leftover dough as this will be the mass of the water and how well we know its sum of MOLS. 1120g - 132g = 988G H2O | H2O = 18g/mol | 988/18 ~ 55 moles H20 FINALLY just calculate the standard formula substituting the values: x = x1/x1 + x2 WE HAVE X = 1 mol + 1 mol SO 55mol X ~ 0.02 * NOTE: it follows the rule that 0 <x <1, also taking the sum of Molara fraction of the solute + water = 1 correctly following the rule concerned Reply
Recommended Articles How the FIES Course Management Accounting Course Law Course Course Civil Engineering Course in Pharmacy Education Course Course Course Psychology of International Relations Distance Learning - Distance Learning Enem Fies Fies Box Jig Enem 2013 EAD Best Undergraduate Universities the 2013 Brazil ProUni Simulated SiSU 2013 Writing Techniques Item Response Theory Career Interest Test Vestibular
Universities of Brazil See list of universities in Brazil with complete information on all universities better paid professions ammonium Learn which are the highest paid and highest-earning professions and opportunities in the current market
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