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A nonvolatile organic compound Z was used to make up a solution. Solution A contains 6.00 g of Zdissolved in 100 g of waterand has a vapor pressure of 754.5 mmHgat the normal boiling point of water. Calculate the molar mass of Z in solution A.

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Answer:

Molar mass of Z in solution A is 148.2 g/mol.

Explanation:

To solve this, we need to use colligative property about vapor pressure.

In any ideal solution, the vapor pressure of solution is lower than vapor pressure of pure solvent. Formula is:

ΔP = P° . Xm

ΔP = Vapor pressure of pure solvent - Vapor pressure of solution

At the normal boiling point of water (100°C), vapor pressure is 760 mmHg.

Let's replace in the formula, to find out Xm (mole fraction of solute)

760 mmHg - 754.5 mmHg = 760 mmHg . Xm

(760 mmHg - 754.5 mmHg) / 760 mmHg = Xm

Xm → 0.00724

Mole fraction of solute = moles of solute /total moles (st + sv)

In this case, our mole fraction will be

(6 g / MM) / ( (6 g/ MM + 100 g / 18g/mol) = 0.00724

Our unknown is MM (molar mass of Z). We solve the equation:

6 g / MM = 0.00724 . ((6 g/ MM + 100 g / 18g/mol))

6 g / MM =  0.04344 g /MM + 0.0402 g/mol

6 g/MM - 0.04344 g /MM = 0.0402 g/mol

5.95656 /MM = 0.0402 g/mol

5.95656 / 0.0402 g/mol = MM → 148.2 g/mol

The molar mass of Z in solution A is 150 g/mol.

Based on the given information,

• The mass of water given is 100 grams.

• The mass of solute given is 6 grams.

• It is known that the molar mass of water is 18 gram per mole.  

Now the moles of water can be calculated as,

[tex]n = \frac{mass}{molar mass}[/tex]

Now putting the values we get,

[tex]n = \frac{100 g}{18 g/mol} \\n = 5.5556 moles[/tex]

• It is known that the vapor pressure of water at 100 degree C (Po) is 760 mmHg.

• The given vapor pressure of solution at 100 degree C is 754.5 mmHg.

• Based on Raoult's law, the vapor pressure of solution = mole fraction of water * Po of water.

The mole fraction of water is,

[tex]= \frac{754.5 mmHg}{760 mmHg} \\= 0.9928[/tex]

The mole fraction of solute is,

[tex]= 1-0.9928\\= 0.00722[/tex]

The mole fraction of solute = n1/n1+n2

Now putting the values we get,

[tex]\frac{n1}{n1+n2} = 0.00722\\\frac{n1+n2}{n1} = \frac{1}{0.00722} \\[/tex]

[tex]\frac{n1+5.56 mol}{n1} = 138.50\\n1 = \frac{5.56 mol}{138.50} \\n1 = 0.040 mol[/tex]

Now the molar mass of solute, Z is,

[tex]Molar mass = \frac{mass}{moles} \\Molar mass = \frac{6.0 grams}{0.04 mol} \\Molar mass = 150 g/mol[/tex]

Thus, the molar mass of Z in solution A is 150 g/mol.

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