Previous Next Progress t welcome kaylin 6. INTRODUCTION Let\'s do a sample calcu

Question

Previous Next Progress t welcome kaylin 6. INTRODUCTION Let's do a sample calculation of this type. A sample of an organic compound that does not dissociate water (i 1.0) weighing 16.5 mg is dissolved in 10.0 mL of water at 24.5 oC on the fixed volume side of an osmometer. The solvent side is filled with distilled water. After the system comes to a steady-state, the osmometer records an osmotic pressure of 224 matm. What is the molar mass of the organic compound? i MRT The expression for the molarity of the compound is M (1.65 x 1013 g Y) (0.0100 L) where Y is the molar mass of the solute. Since we know i, IT, R and T (remember to convert the temperature to Kelvin), we can solve for Y. Set this up and solve for Y; then, enter the result below (use 2 sig figs in your answer). g/mol The organic compound in this example was glucose, C.Hizo.

Explanation / Answer

Given

pi = i x M x R x T

also given

pi= 224 x 10-3 atm

i = 1

R = 0.0821

T = 24.5 C

T = 24.5 + 273 K

T = 297.5 K

so

using those values

we get

pi = i x M x R x T

224 x 10-3 = 1 x M x 0.0821 x 297.5

M = 9.171 x 10-3

we know that

molarity = moles / volume (L)

also

moles = mass / molar mass

given

mass = 1.65 x 10-2

volume = 0.01 L

so

molarity = (1.65 x 10-2 / Y) / 0.01

where Y is the molar mass

Now

the given expression for molarity is

M = ( 1.65 x 10-2 / Y) / (0.01)

substitute the value of M in this equation we get

9.171 x 10-3 = ( 1.65 x 10-2 /Y) / 0.01

we get

Y = 179.91

so the molar mass of glucose is 179.91 g

rounding to two siginificant figures

we get
the molar mass of glucose is 180 g

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