Question

An isotonic solution will produce an osmotic pressure of $10 \cdot 00{\text{ atm}}$ measured against pure water at ${37^ \circ }{\text{C}}$. How many grams of ${\text{NaCl}}$ must be dissolved in one litre of water to produce isotonic solution?
A. $11.46{\text{ g}}$
B. $0.196{\text{ g}}$
C. $9.01{\text{ g}}$
D. $10{\text{ g}}$

Answer 1

The solutions having the same concentration of salts are known as isotonic solutions. As the concentrations are the same, the isotonic solutions have the same osmotic pressure.
The pressure applied to pure solvent to prevent the solvent from passing into the solution is known as osmotic pressure. The osmotic pressure is calculated using the equation,
$\pi = iMRT$
Where, $\pi$ is the osmotic pressure,
$i$ is the van’t Hoff factor,
$M$ is the molar concentration of the solution,
$R$ is the universal gas constant,
$T$ is the temperature.

Complete step by step answer:
Step 1: Calculate the molar mass of ${\text{NaCl}}$ as follows:
${\text{Molar mass of NaCl}} = \left( {1 \times {\text{Mass of Na}}} \right) + \left( {1 \times {\text{Mass of Cl}}} \right)$
$= \left( {1 \times 22 \cdot 99} \right) + \left( {1 \times 35 \cdot 45} \right)$
$= 22 \cdot 99 + 35 \cdot 45$
${\text{Molar mass of NaCl}} = 58 \cdot 44{\text{ g mo}}{{\text{l}}^{ - 1}}$
Thus, the molar mass of ${\text{NaCl}}$ is $58 \cdot 44{\text{ g mo}}{{\text{l}}^{ - 1}}$
Step 2: Convert the units of temperature from $^ \circ {\text{C}}$ to ${\text{K}}$ using the relation as follows:
$T\left( {\text{K}} \right) = {T^ \circ }{\text{C}} + 273$.
Substitute ${37^ \circ }{\text{C}}$ for the temperature in $^ \circ {\text{C}}$. Thus,
$T\left( {\text{K}} \right) = {37^ \circ }{\text{C}} + 273$
$T\left( {\text{K}} \right) = 310{\text{ K}}$
Thus, the temperature is $310{\text{ K}}$.
Step 3: Calculate the molar concentration of the solution as follows:
The molar concentration of the solution is the number of moles of solute dissolved in one litre of solvent. Thus,
${\text{M}} = \dfrac{{{\text{Number of moles of NaCl}}\left( {{\text{mol}}} \right)}}{{1{\text{ L}}}}$
The number of moles of any solute is the ratio of mass in grams to the molar mass of the solute. Thus,
${\text{M}} = \dfrac{{{\text{Mass of NaCl}}\left( {\text{g}} \right)/{\text{Molar mass of NaCl}}\left( {{\text{g mo}}{{\text{l}}^{ - 1}}} \right)}}{{1{\text{ L}}}}$
Let the mass in grams of ${\text{NaCl}}$ be $x$. The molar mass of ${\text{NaCl}}$ is $58 \cdot 44{\text{ g mo}}{{\text{l}}^{ - 1}}$. Thus, the molar concentration of the solution is,
${\text{M}} = \dfrac{{x{\text{ g}}/58 \cdot 44{\text{ mo}}{{\text{l}}^{ - 1}}}}{{1{\text{ L}}}}$
${\text{M}} = \dfrac{{x{\text{ g}}}}{{58 \cdot 44{\text{ mo}}{{\text{l}}^{ - 1}}{\text{ L}}}}$
Thus, the molar concentration of the solution is $\dfrac{{x{\text{ g}}}}{{58 \cdot 44{\text{ mo}}{{\text{l}}^{ - 1}}{\text{ L}}}}$.
Step 4: Calculate the grams of ${\text{NaCl}}$ must be dissolved in one litre of water to produce isotonic solution as follows:
The osmotic pressure is calculated using the equation,
$\pi = iMRT$
Rearrange the equation for the molar concentration. Thus,
$M = \dfrac{\pi }{{iRT}}$
Substitute $\dfrac{{x{\text{ g}}}}{{58 \cdot 44{\text{ mo}}{{\text{l}}^{ - 1}}{\text{ L}}}}$ for the molar concentration, $10 \cdot 00{\text{ atm}}$ for the osmotic pressure, $2$ for the van’t Hoff factor of ${\text{NaCl}}$, $0 \cdot 08206{\text{ L atm mo}}{{\text{l}}^{ - 1}}{\text{ }}{{\text{K}}^{ - 1}}$ for the universal gas constant and $310{\text{ K}}$ for the temperature. Thus,
$\dfrac{{x{\text{ g}}}}{{58.44{\text{ }}\not{{{\text{mo}}{{\text{l}}^{ - 1}}}}\not{{\text{L}}}}} = \dfrac{{10 \cdot 00{\text{ }}\not{{{\text{atm}}}}}}{{2 \times 0 \cdot 08206{\text{ }}\not{{\text{L}}}{\text{ }}\not{{{\text{atm}}}}{\text{ }}\not{{{\text{mo}}{{\text{l}}^{ - 1}}}}{\text{ }}\not{{{{\text{K}}^{ - 1}}}} \times 310{\text{ }}\not{{\text{K}}}}}$
$x = 11 \cdot 46{\text{ g}}$
Thus, the grams of ${\text{NaCl}}$ must be dissolved in one litre of water to produce isotonic solution $11 \cdot 46{\text{ g}}$.

So, the correct answer is Option A .

Note:
The ratio of actual concentration of particles produced on dissolving a substance to the concentration of substance calculated from its mass is known as van’t Hoff factor.
${\text{NaCl}}$ gives two ions on dissociation, ${\text{N}}{{\text{a}}^ + }$ and ${\text{C}}{{\text{l}}^ - }$.Thus, the van’t Hoff factor for ${\text{NaCl}}$ is $2$.