What osmotic pressure in mm Hg would you expect for an aqueous so...

Question

What osmotic pressure in mm Hg would you expect for an aqueous solution of 11.5 mg of insulin 1mol. weight = 59902 
in 6.60 mL of solution at 298 K? What would the height of the water column be in meters? The density of mercury is 
13.534 g/mL at 298 K.

Accepted Answer

well, everyone's on this problem with calculation for the automatic pressure in millimeters of mercury for an agreed solution are also being asked what the water columns, high beam and meters were also given the density of mercury to work with the equation that we're going to use is the automatic pressure is equal to capital M R. T. So this pie here, that's just static pressure. Capital M here stands for my clarity. The capital R is our ideal gas constant. So let's write this out that 0.8 to 06 units being leaders times a tm over kelvin's times moles. So all the units here in this equation should match with the ideal ideal gas law constant. So capital T then is going to be temperature of course, units being Calvin's. So now we do have all the necessary information to solve for r osmotic pressure. So for clarity, we know that clarity is moles of our sauce. Over. The leaders of our solution were only given the milligrams and the male leaders. So we need to go ahead and do some dimensional analysis within this equation. So starting off with the moles. Then again, we're given the 9.31 mg of listen up in. We go ahead and convert the milligrams into grams. So for every 1000 mg we have one g. And then using the given molar mass, we can convert our grams into moles. So, well just about everything here. So for every one mole of let's open, we have 904 500 sorry, 537 g. So that's our mole. Computer components of our polarity. Then on the bottom we need the leaders of solution were given 5.12 mL will convert this into leaders. So for every 1000 mL we have one leader. So we see here that the units of milligram will cancel as well as grams and milliliters will cancel here. So that here is just all for capital M. Now for capital, are we already brought this out? This is 0.8206 units being layers times A T. M. For every kelvin's times mold, You know, for capital T temperature which you've already given to us in Calvin's, which is um 298 Calvin's. So now we see here with multiplying our capital are the ideal gas constant. Our moles will cancel. Our leaders will cancel and our kelvin's will cancel. Leaving us with the osmotic pressure being equal to 0.8280 A. T. M. But in this problem we're being asked for the automatic pressure in units of millimeters of mercury. So what grade correct conversion here? So for every 1 80 M we have millimeters of mercury. You see here now that the units of 80 M will cancel. So then the numerical value that we get. Once we put everything into a calculator we will get 62.9313 millimeters of mercury. And last week actually for the Hi ever column. So that's equal to starting off with the osmotic pressure being 62.9313 millimeters of mercury. We're gonna go ahead and use the density here. So we have 100 or 13.534 M. M. Of H 20 with one mm mm of mercury. That will do for everyone. M. We have 1000 m.m. We see here that this will cancel as well as this right here, Leaving us with the height being 0.8517. So this right here is going to be my final answer for this problem.

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