**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 Â° 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**.

## Expert Answer

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

\[Â \amalg = M R T \]

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**.

\[Â \amalg = M R T \]

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:

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

\[M = 0. 0131 mol L ^ -1 \]

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**.*