Question

In: Chemistry

A company is diluting concentrated 80% glycerol and diluting it to 6 wt% glycerol with water...

A company is diluting concentrated 80% glycerol and diluting it to 6 wt% glycerol with water to produce for use in the pharmaceutical industry.

(a) What are the average molecular weights of the starting and product glycerol solutions?

(b) What are the mole fractions of the initial and final glycerol solutions? (Assume the density of water is 1 g/cm3 and the specific gravity of glycerol is 1.26. )

Expert Solution

(a) Average molar mass = x_wM_w + x_gM_g

x_w and M_w = mole fraction and Molecular mass of water

x_g and M_g = mole fraction and Molecular mass of water

(i) starting solution

80 wt% glycerol solution means 80g of glycerol are present in 100g of water

Malarity of glycerol = No. of moles / Volume of solution in liters

No. of moles of glycerol = mass / molecular mass = 80/92 = 0.87 moles

Volume of solution = Volume of water + Volume of glycerol

Volume = mass / density

Volume of solution = 100/1 + 80/1.26 = 163.5mL = 0.163L

Molarity of glycerol = 0.87 / 0.163 = 5.34M

Molarity of water = (100/18 X 0.163) = 34.08

Mole fraction of glycerol x_g = No. of moles of glycerol / total no. of moles = 5.34 / (5.34 + 34.08)

Mole fraction of glycerol x_g = 0.135

Mole fraction of water x_w= 1 - 0.135 =0.865

Average molar mass = 0.865 X 18 + 0.135 X 92.0 = 27.99

(ii) after dilution

6g of glyserol are present in 100g of water

Total volume = 6/1.26 + 94/1 = 98.76mL = 0.0988L

Molarity of glycerol = 6 / (92.0 X 0.0988) = 0.665M

Molarity of water = 94 / (18 X 0.0988) =52.86M

Mole fraction of glycerol x_g = 0.665 / (0.665 + 52.86) = 0.0124

Mole fraction of water x_w = 52.86 / (52.86 + 0.665) = 0.987

Average molar mass = 0.987 X 18 + 0.0124 X 92.0 = 18.90

(b) The initial and final mole fractions of glycerol are calculated above.

queen_honey_blossom answered 2 years ago

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