276 grams of carbon in 23.0 moles
Answer: 0.050M urea, 0.10M glucose, 0.2M sucrose, pure water
Explanation:
Vapor pressure refers to the ease with which a liquid substance is transformed into vapour. High vapour density implies that the liquid is easily transformed into gas. Pure water is expected to have the lowest vapour density since it is held by strong intermolecular forces in the liquid state. Urea is an organic liquid held by weak Van der Waals forces hence its extremely high vapor pressure.
Toulene = 35.6 g
Benzene = 125 g = 0.125 kg
Molecular weight of Toluene C6H5CH3 = 92.1g/mol
Moles of toulene = 35.6 g / 92.1 g/mol = 0.3865 mol
Now the molarity of the toulene in the given solution = 0.3865 / 0.125 = 3.092 m
Molarity of C6H5CH3 = 3.092 m
The balanced reaction is 2KClO3 --> 2KCl + 3O2
We first divide the 400.0 g KClO3 by the molar mass of 122.55 g/mol to get 3.26 mol KClO3. Next, we use the coefficients: 3.26 mol KClO3 * (3 mol O2 / 2 mol KClO3) = 4.896 mol O2. Multiplying this by the molar mass of 32 g/mol gives 156.67 g O2.
Percent yield = 115.0 g / 156.67 g = 0.734 = 73.40%
