An aqueous solution containing 1.00 g of bovine insulin (a protein, not ionized) per liter has an osmotic pressure of 3.1 mm Hg at 25 °C. How do you calculate the molar mass of bovine insulin?

Answer 1

The molar mass is #6.0 ×10^3color(white)(l)"g/mol"#.

The formula for osmotic pressure #Pi# is
#color(blue)(bar(ul(|color(white)(a/a)Pi = cRTcolor(white)(a/a)|)))" "#

where

#c =# the molar concentration of the solute #R =# the universal gas constant #T =# the Kelvin temperature of the solution
Since #c = "moles"/"litres" = n/V#, we can write
#Pi = (nRT)/V#
Since #n ="mass"/"molar mass"= m/M#, we can write
#Pi = (nRT)/(MV)#

We can solve this equation for the molar mass and get

#color(blue)(bar(ul(|color(white)(a/a)M=(mRT)/(PiV)color(white)(a/a)|)))" "#

In this problem

#m = "1.00 g"# #R = "0.082 06 L·atm·K"^"-1""mol"^"-1"# #T = "25 °C = 298.15 K"# #Pi = 3.1 color(red)(cancel(color(black)("mmHg"))) × "1 atm"/(760 color(red)(cancel(color(black)("mmHg")))) = "0.004 08 atm"# #V = "1 L"#
∴ #M = ("1.00 g" × "0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1")))"mol"^"-1" × 298.15 color(red)(cancel(color(black)("K"))))/("0.00 408"color(red)(cancel(color(black)("atm"))) × 1 color(red)(cancel(color(black)("L")))) = 6.0 × 10^3color(white)(l)"g/mol"#
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Answer 2

To calculate the molar mass of bovine insulin, you can use the formula for osmotic pressure:

π = MRT

Where: π = osmotic pressure M = molarity of the solution (in mol/L) R = ideal gas constant (0.0821 L atm/mol K) T = temperature (in Kelvin)

Rearrange the formula to solve for molarity (M):

M = π / RT

Given: π = 3.1 mm Hg = 3.1 × 10^(-3) atm (since 1 mm Hg = 1 torr ≈ 1 atm) R = 0.0821 L atm/mol K T = 25 °C = 298 K

Substitute the values:

M = (3.1 × 10^(-3) atm) / (0.0821 L atm/mol K × 298 K)

Then, solve for molarity (M).

Once you have the molarity (M), you can calculate the number of moles of bovine insulin using the given mass and volume of the solution. Finally, divide the mass of bovine insulin by the number of moles to find the molar mass.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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