 # In a study designed to prepare new gasoline.

This question aims to find the molar mass of the polymer sample when a study is designed to prepare gasoline-resistant coatings.

A polymer engineer was making gasoline-resistant coatings. To make these new coatings, the engineer dissolved 6.053 g of polyvinyl alcohol in water to make 100.0 mL of the solution. The osmotic pressure of the solution of polyvinyl alcohol is 0.272 atm at 25 ° C.

The minimum pressure that can be applied to the solution to prevent the flow of solvent through a semi-permeable membrane is called osmotic pressure. The osmotic pressure highly depends on the size of dissolved particles in the solution. It is represented by $\amalg$ and its unit is atm.

The molarity of this solution is calculated by the following formula:

$\amalg = M R T$

Here, M represents the molarity, $\amalg$ represents the osmotic pressure, T stands for temperature, R and K represents the gas constant. Molarity is the concentration of solute in the specific volume of the solution.

The equation of gas constant is:

$R = 0 . 8 2 1 \frac { atm \times L } { mol \times K }$

Rearranging the equation of osmotic pressure to get molarity:

$M = \frac {\amalg} {RT}$

By putting the values in the expression:

$M = \frac { 0.272} { 0.0821 atm L mol ^ – 1 K^- 1 \times 298 . 15 K}$

$M = 0. 011 mol L ^ -1$

We can calculate the moles n from the following formula:

$M = \frac { n _ { solute } } { V _ { solution } }$

$n _ { solute } = M \times V _ { solution }$

$n _ { solute } = 0. 011 mol L^-1 \times 100 \times 10^-3 L$

$n _ { solute } = 1.1 \times 10 ^ -3 mol$

The molar mass of the solution is calculated by:

$n = \frac { mass } { molar mass }$

$M = \frac { m } { n }$

$M = \frac { 6 . 053 g } { 1.1 \times 10 ^ -3 mol }$

$M = 5502 . 73 g/mol$

## Numerical Solution

The molar mass of the polymer sample is 5502 . 73 g/mol.

## Example

Consider a polymer engineer making a coating having osmotic pressure 0.321 atm with the same parameters as mentioned above. Find the molar mass of the polymer sample.

$\amalg = M R T$

The equation of gas constant is:

$R = 0.821\frac {atm \times L } { mol \times K }$

Rearranging the equation of osmotic pressure to get molarity:

$M = \frac {\amalg} { RT}$

By putting the values in the expression:

$M = \frac { 0 . 3 2 1 } { 0 . 0 8 2 1 atm L mol ^ – 1 K ^ – 1 \times 298 . 15 K }$

$M = 0. 0131 mol L ^ -1$

We can calculate the moles n from the following formula:

$M = \frac { n _ { solute } } { V _ { solution } }$

$n _ { solute } = M \times V _ { solution }$

$n _ { solute } = 0. 0131 mol L ^ -1 \times 100 \times 10 ^ -3 L$

$n _ { solute } = 1.31 \times 10 ^ -3 mol$

The molar mass of the solution is calculated by:

$n = \frac { mass } { molar mass }$

$M = \frac {m}{n}$

$M = \frac {6 . 053 g } { 1.31 \times 10 ^ -3 mol }$

$M = 4620 . 61 g/mol$

Image/Mathematical drawings are created in Geogebra.

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