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3 years ago

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What is the vapor pressure (in mm Hg) of a solution of 17.5 g of glucose (C6H12O6) in 82.0 g of methanol (CH3OH) at 27∘C? The va

por pressure of pure methanol at 27∘C is 1.40×102 mm H

1 answer:

vichka [17]3 years ago

5 0

Answer:

134.8 mmHg is the vapor pressure for solution

Explanation:

We must apply the colligative property of lowering vapor pressure, which formula is: P° - P' = P° . Xm

P° → Vapor pressure of pure solvent

P' → Vapor pressure of solution

Xm → Mole fraction for solute

Let's determine the moles of solute and solvent

17.5 g . 1 mol/180 g = 0.0972 moles

82 g . 1mol / 32 g = 2.56 moles

Total moles → moles of solute + moles of solvent → 2.56 + 0.0972 = 2.6572 moles

Xm → moles of solute / total moles = 0.0972 / 2.6572 = 0.0365

We replace the data in the formula

140 mmHg - P' = 140 mmHg . 0.0365

P' = - (140 mmHg . 0.0365 - 140mmHg)

P' = 134.8 mmHg

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Read 2 more answers

What is the concentration of a solution containing 1.11 g sugar (sucrose, C12H22O11, MW = 342.3 g/mol, d = 1.587 g/cm3) in 432 m

djyliett [7]

Answer:

0.0075 M

0.0060 m

Explanation:

Our strategy here is to use the definition of molarity and molality to solve this question.

The molarity is the number of moles of solute, sucrose in this case, per liter of solution.

The molality is the number of moles of solute per kilogram of solvent.

So the molarity of the solution is

M = moles of solute/ V solution

As we see we need the volume of solution since we are only given the volume of solvent, but this will be easy to compute since we have the density of sucrose.

So determine the moles of sucrose , and the volume of solution:

Moles sucrose = 1.11 g/342.3 g/mol = 3.24 x 10⁻³ M

Volume of solution = Vol Sucrose + Vol glycerine

d = m/V ⇒ Vsucrose = m / d = 1.11 g/ 1.587 g/cm³ = 0.70 cm³

Vol solution = 432 mL + 0.70 mL = 432.7 mL (1cm³ = 1 mL)

Vol solution = 432.7 mL x 1 L / 1000 mL = 0.4327 L

⇒ M = 3.2 x 10⁻³ mol / 0.4327 L = 0.0075 M

For the molarity what we need is to first calculate the kilograms of glycerine from the given density:

d = m/v ⇒ m = d x v = 1.261 g/cm³ x 432 cm³ = 544.75 g

Converting to Kg:

544.75 g x 1 Kg/ 1000 g = 0.544 kg

Now the molality is

m = mol sucrose/ kg solvent = 3.24 x 10⁻³ mol / 0.544 Kg = 0.0060 m

Note: In the calculation for volume of solution we could have approximated it to that of just glycerine, but since the density of sucrose was given we calculated the total volume of solution to be more rigorous.

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