Question

classify the tonicity of a solution made by dissolving 58 g of sodium chloride in 9.7 l of water.
choose the correct answer and explanation below.
select the correct answer below:
the solution is hypertonic because it has a 58% m/v concentration of sodium chloride.
the solution is hypotonic because it has a 58% m/v concentration of sodium chloride.
the solution is isotonic because it has a 58% m/v concentration of sodium chloride.
the solution is hypertonic because it has a 0.60% m/v concentration of sodium chloride.
the solution is hypotonic because it has a 0.60% m/v concentration of sodium chloride.
the solution is isotonic because it has a 0.60% m/v concentration of sodium chloride.

Explanation:

Step1: Calculate the mass/volume percentage

The formula for mass/volume percentage ($\% \text{m/v}$) is $\frac{\text{mass of solute (g)}}{\text{volume of solution (mL)}}\times100$.
First, convert the volume of water (assuming the volume of the solution is approximately equal to the volume of water) from liters to milliliters. Since $1\ \text{L} = 1000\ \text{mL}$, $9.7\ \text{L}=9.7\times1000 = 9700\ \text{mL}$.
The mass of sodium chloride ($\text{NaCl}$) is $m = 58\ \text{g}$.
Then, $\% \text{m/v}=\frac{58\ \text{g}}{9700\ \text{mL}}\times 100\approx0.60\%$.

Step2: Determine the tonicity

An isotonic solution has a $\text{NaCl}$ concentration of approximately $0.9\%$. A hypotonic solution has a lower $\text{NaCl}$ concentration than the cell's interior (less than $0.9\%$), and a hypertonic solution has a higher $\text{NaCl}$ concentration (greater than $0.9\%$).
Since $0.60\%<0.9\%$, the solution is hypotonic.

Answer:

The solution is hypotonic because it has a $0.60\%\ \text{m/v}$ concentration of sodium chloride.