Molarity Review Molarity Stoichiometry Problems Youtube A review of how to complete a stoichiometry problem using molarity. consider watching this video after you've seen this material in lecture, or after you've. This example shows three different types of ways a solution stoichiometry question can be asked, using molarity, stoichiometry and dilutions. i walk you thro.
Stoichiometry Lesson 3 Molarity Problems Youtube This chemistry video tutorial explains how to solve solution stoichiometry problems. it discusses how to balance precipitation reactions and how to calculat. Problem #2: what is the molarity of 245.0 g of h 2 so 4 dissolved in 1.000 l of solution? solution: mv = grams molar mass (x) (1.000 l) = 245.0 g 98.0768 g mol¯ 1. x = 2.49804235 m to four sig figs, 2.498 m if the volume had been specified as 1.00 l (as it often is in problems like this), the answer would have been 2.50 m, not 2.5 m. Problem 6.1.5 6.1. 5. calculate the number of moles and the mass of the solute in each of the following solutions: (a) 2.00 l of 18.5 m h 2 so 4, concentrated sulfuric acid. (b) 100.0 ml of 3.8 × 10 −5 m nacn, the minimum lethal concentration of sodium cyanide in blood serum. (c) 5.50 l of 13.3 m h 2 co, the formaldehyde used to “fix. The molar mass of cocl 2 •2h 2 o is 165.87 g mol. therefore, molescocl2 ⋅ 2h2o = (10.0g 165.87g mol) = 0.0603mol. the volume of the solution in liters is. volume = 500 ml(1l 1000ml) = 0.500l. molarity is the number of moles of solute per liter of solution, so the molarity of the solution is.
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