Osmotic Pressure Example 1

Jules Bruno
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here, it says. Calculate the osmotic pressure of the solution that is 18.30 mg of zinc oxide in 15.1 ml of solution at 26 degrees Celsius. Alright, so osmotic pressure equals I times polarity times r times T zinc oxide is an ionic solute that breaks up into zinc ion and oxide ion. That's two ions. So I equals two capital M is our polarity, which is moles over liters here. When I convert the 15 MLS into liters, that's 150.15 liters. Then I can change the 8.30 mg of zinc into moles. So remember, one mg is 10 to the negative three g and one mole of zinc, the weight of zinc oxide, the mole, the weight of zinc oxide. When we figure that out is Well, the masses 81 3, 8 g. So when we work that out, the moles is 2.2487 times 10 to the - moles. So then that's gonna give us our moles over liters or is R gas constant which we don't have to do anything. We just have to plug it in, And then our temperature needs to be in Kelvin. So add to 73.15 to this number here, and that gives us to 99.15 Kelvin. So then what cancels out Kelvin's? Cancel out moles, cancel out leaders, cancel out and we're left with atmospheres at the end. So we plug this in, we're gonna get as our atmosphere's . atmospheres as the osmotic pressure for this given solution.