The osmotic pressure of blood is 5950.8 mmHg at 41°C. What mass of glucose, C 6 H 12 O 6 , is needed to prepare 5.51 L of solution. The osmotic pressure of the glucose solution is equal to the osmotic pressure of blood.

We're told the osmotic pressure of blood is 5,950.8 millimeters of mercury at 41 degrees Celsius. What mass of glucose, which is c 6h1206, is needed to prepare 5.51 liters of solution? The osmotic pressure of the glucose solution is equal to the osmotic pressure of blood. Alright. So osmotic pressure equals I times molarity times r times t. What we need to realize is osmotic pressure is going to be in atmospheres, so what I'm gonna do first is I'm gonna convert the millimeters of mercury into atmospheres. So one atmosphere is 760 millimeters of mercury. So when we do that we get 7.83 atmospheres here as our osmotic pressure. And that's gonna equal 1 because our solid of glucose is covalent. Covalent things don't break up into ions. Here, we don't know the moles of glucose, that's what we need to find, divided by liters of solution, which is given to us. R is our gas constant, and t is our temperature in Kelvin. So I'm gonna add 273.15 to this number here and that'll give me my Kelvin. So that's 314.15 Kelvin. All we have to do now is solve for x. So I'm gonna multiply r times this temperature. So when I do that I'm going to get 7.83 equals 25.779149 times x divided by 5.51. Again, where did I get this 25 in change? Came from multiplying r times the temperature. Now cross multiply 7.83 times 5.51, when I do that I'm gonna get 43.14 33 and that still equals 25.779149x, divide both sides now by 43.1433 actually, not by that, by 25 0.779149, and that'll give me my x. So we're gonna work it over here. So here your x equals 1 point 67357 moles of glucose. Because remember our x that we're solving for was our moles of glucose. If I know the moles, then I can figure out the the mass. We just say 1 mole of glucose and its molecular weight or molar mass is 180.156 grams. That's the weight of the 6 carbons, 12 hydrogens, and 6 oxygens. So that gives me as my total at the end, 302 grams of glucose. So that would be my final answer.

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14. Solutions

Osmotic Pressure

General Chemistry

14. Solutions

Osmotic Pressure

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Multiple Choice

The osmotic pressure of blood is 5950.8 mmHg at 41°C. What mass of glucose, C₆H₁₂O₆, is needed to prepare 5.51 L of solution. The osmotic pressure of the glucose solution is equal to the osmotic pressure of blood.

54.7 g

0.304 g

419 g

302 g

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Verified step by step guidance

First, understand that osmotic pressure (π) can be calculated using the formula: π = iMRT, where i is the van't Hoff factor (which is 1 for glucose, a non-electrolyte), M is the molarity of the solution, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin.

Convert the given temperature from Celsius to Kelvin by adding 273.15 to the Celsius temperature: T(K) = 41 + 273.15.

Convert the osmotic pressure from mmHg to atm using the conversion factor: 1 atm = 760 mmHg. So, π(atm) = 5950.8 mmHg / 760 mmHg/atm.

Rearrange the osmotic pressure formula to solve for molarity (M): M = π / (iRT). Substitute the values for π, i, R, and T into the equation to find the molarity of the glucose solution.

Finally, calculate the mass of glucose needed using the formula: mass = Molarity (M) × Volume (L) × Molar Mass of glucose (C6H12O6, which is approximately 180.18 g/mol). Use the volume of the solution (5.51 L) and the molarity calculated in the previous step to find the mass of glucose required.