Question

Isotonic saline solution, which has the same osmotic pressure as blood, can be prepared by dissolving 0.923 g of nacl in enough water to produce 100. ml of solution. what is the osmotic pressure of this solution at 25 ∘c

Answer 1

osmotic pressure = CRT to get C, get the number of moles of NaCl then divide it with volume of the solution mole NaCl = 0.923g / 58.442 g/mol mole NaCl = 0.0158 mol C = 0.0158 mol / 0.1L C = 0.158 mol/L So, osmotic pressure = 0.158 mol/L*(0.082058 L-atm/mol-K)*(25+273K) osmotic pressure = 3.86 atm

Answer 2

The osmotic pressure of NaCl solution at $25\;^\circ{\text{C}}$ is $\boxed{{\text{3}}{\text{.86 atm}}}$.

Further Explanation:

Osmotic Pressure:

It is the measure of the tendency of a solution to take in pure solvent via osmosis. The minimum pressure that is to be applied to the solution to prevent the inward flow of the pure solvent across the semipermeable membrane. Osmosis occurs when two solutions have different concentrations of solute and are separated by a semipermeable membrane.

The formula to calculate the osmotic pressure of the solution is as follows:

$\prod={\text{MRT}}$                                                                 …… (1)

Here, $\prod$ is the osmotic pressure.

M is the molarity of the solution.

R is universal gas constant.

T is the absolute temperature.

Molarity is a concentration term that is defined as the number of moles of solute dissolved in one litre of the solution. It is denoted by M and its unit is mol/L.

The formula to calculate the molarity of NaCl is as follows:

${\text{Molarity of NaCl solution}}=\frac{{{\text{amount}}\;\left({{\text{mol}}}\right)\;{\text{of}}\;{\text{NaCl}}}}{{\;{\text{volume }}\left({\text{L}}\right)\;{\text{of NaCl}}\;{\text{solution}}}}$        ...... (2)

The amount of NaCl is calculated as follows:

${\text{Amount of NaCl}}=\frac{{{\text{Given mass of NaCl}}}}{{{\text{Molar mass of NaCl}}}}$              …… (3)

The given mass of NaCl is 0.923 g.

The molar mass of NaCl is 58.442 g/mol.

Substitute these values in equation (3).

$\begin{gathered}{\text{Amount of NaCl}}=\left({{\text{0}}{\text{.923 g}}}\right)\left({\frac{{{\text{1 mol}}}}{{{\text{58}}{\text{.442 g}}}}}\right)\\={\text{0}}{\text{.01579 mol}}\\\end{gathered}$

The volume of NaCl is to be converted in L. The conversion factor for this is,

 ${\text{1 mL}}={10^{-3}}\;{\text{L}}$

So the volume of NaOH is calculated as follows:

$\begin{gathered}{\text{Volume of NaCl}}=\left({{\text{100 mL}}}\right)\left({\frac{{{{10}^{-3}}\;{\text{L}}}}{{{\text{1 mL}}}}}\right)\\=0.{\text{1 L}}\\\end{gathered}$

The amount of NaCl is 0.01579 mol.

The volume of NaCl is 0.1 L.

Substitute these values in equation (2).

$\begin{gathered}{\text{Molarity of NaCl solution}}=\frac{{{\text{0}}{\text{.01579 mol}}}}{{{\text{0}}{\text{.1 L}}}}\\={\text{0}}{\text{.1579 M}}\\\end{gathered}$

The temperature is to be converted into K. The conversion factor for this is,

${\text{0}}^\circ{\text{C}}={\text{273 K}}$

So the temperature of the solution is calculated as follows:

$\begin{gathered}{\text{Temperature}}\left({\text{K}}\right)=\left({25 + 273}\right)\;{\text{K}}\\=298\;{\text{K}}\\\end{gathered}$

The value of M is 0.1579 M.

The value of R is ${\text{0}}{\text{.0821}}\;{\text{L}}\cdot{\text{atm/mol}}\cdot {\text{K}}$.

The value of T is 298 K.

Substitute these values in equation (1).

$\begin{gathered}\prod=\left({{\text{0}}{\text{.1579 M}}}\right)\left({{\text{0}}{\text{.0821}}\;{\text{L}}\cdot{\text{atm/mol}}\cdot{\text{K}}}\right)\left({{\text{298 K}}}\right)\\={\text{3}}{\text{.86314}}\;{\text{atm}}\\\approx{\text{3}}{\text{.86 atm}}\\\end{gathered}$

So the osmotic pressure of NaCl solution at ${\mathbf{25}}\;{\mathbf{^\circ C}}$ is 3.86 atm.

Learn more:

1. Which statement is true for Boyle’s law: brainly.com/question/1158880

2. Calculation of volume of gas: brainly.com/question/3636135

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Colligative properties

Keywords: osmotic pressure, molarity, amount of NaCl, 3.86 atm, 298 K, 0.1579 M, molar mass of NaCl, given mass of NaCl, R, T, M, universal gas constant, absolute temperature.