Isotonic and buffer solutions: practlice questions
Adfusting toniclly:
1. How much of NaCl should be added to $200 \mathrm{ml}$ of an aqueous solution confalinhe $0.5 \%$ Atropin Sulfate to make il isotonic? How much of dextrose in place of Naci choutd be used in order to make this solution lsotonke? You are given the follow Informatlon.
\begin{tabular}{lc}
Compound & Evalue \
Atropln Sullate & 0.13 \
NaCl & 1.00 \
Dextrose & 0.16
\end{tabular}

Question

Isotonic and buffer solutions: practlice questions
Adfusting toniclly:
1. How much of NaCl should be added to $200 \mathrm{ml}$ of an aqueous solution confalinhe $0.5 \%$ Atropin Sulfate to make il isotonic? How much of dextrose in place of Naci choutd be used in order to make this solution lsotonke? You are given the follow Informatlon.
\begin{tabular}{lc} 
Compound & Evalue \
Atropln Sullate & 0.13 \
NaCl & 1.00 \
Dextrose & 0.16
\end{tabular}

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Calculate the total E value needed for isotonicity**: To make a solution isotonic, the total E value of the solution should be equivalent to that of a 0.9% NaCl solution, which has an E value of 1.00. ‖ ‖ 2 ⋮ **Determine the E value contribution of Atropin Sulfate**: The E value for Atropin Sulfate is given as 0.13. Since the solution contains 0.5% Atropin Sulfate, we calculate its contribution to the E value by multiplying the percentage by the E value: $0.5\% imes 0.13 = 0.065$. ‖ ‖ 3 ⋮ **Calculate the E value deficit**: Subtract the E value contribution of Atropin Sulfate from the total E value needed for isotonicity: $1.00 - 0.065 = 0.935$. This is the E value deficit that needs to be met by adding NaCl or dextrose. ‖ ‖ 4 ⋮ **Calculate the amount of NaCl needed**: To find the amount of NaCl needed to make up the E value deficit, divide the E value deficit by the E value of NaCl: $0.935 \div 1.00 = 0.935\%$. To find the mass of NaCl needed for the 200 ml solution, convert the percentage to grams: $0.935\% imes 200\, \mathrm{ml} imes \frac{1\, \mathrm{g}}{100\, \mathrm{ml}} = 1.87\, \mathrm{g}$. ‖ ‖ 5 ⋮ **Calculate the amount of dextrose needed**: To find the amount of dextrose needed to make up the E value deficit, divide the E value deficit by the E value of dextrose: $0.935 \div 0.16 \approx 5.844\%$. To find the mass of dextrose needed for the 200 ml solution, convert the percentage to grams: $5.844\% imes 200\, \mathrm{ml} imes \frac{1\, \mathrm{g}}{100\, \mathrm{ml}} = 11.688\, \mathrm{g}$. ‖ ∻1 Answer∻ ⚹ To make the solution isotonic, add 1.87 g of NaCl or 11.688 g of dextrose. ⚹ ∻Key Concept∻ ⚹ Isotonic solutions have the same osmotic pressure as body fluids, and the E value is used to adjust the tonicity of solutions. ⚹ ∻Explanation∻ ⚹ The E value represents the equivalent amount of NaCl that has the same osmotic effect as 1% of the compound in question. By using the E values, we can calculate the amount of NaCl or dextrose needed to make a solution isotonic. ⚹◊What is the relationship between the E values of compounds and their ability to adjust tonicity in a solution⍭ Generate me a similar question◊

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