# Question #daedf

Sep 5, 2017

The osmolality of the solution is 304 mOsmol/kg.

#### Explanation:

An osmole (Osmol) is 1 mol of particles that contribute to the osmotic pressure of a solution.

The problem here is how to convert equivalents to osmoles.

The point to remember is that 1 eq of anything reacts with or produces 1 eq of anything else

For example,

$\text{CaCl"_2 → "Ca"^"2+" + "2Cl"^"-}$
$\text{1 mol"color(white)(mm)"1 mol"color(white)(ml)"2 mol}$
$\text{2 eq"color(white)(mmll)"2 eq"color(white)(mml)"2 eq}$

We see that $\text{2 mol Cl"^"-" = "2 eq Cl"^"-}$, but $\text{1 mol Ca"^"2+" = "2 eq Ca"^"2+}$.

For ions, the rule is, $\text{1 mol" = ncolor(white)(l) "eq}$ or

$\boldsymbol{\text{1 eq" = 1/n color(white)(l)"mol}}$,

where $n$ is the absolute value of the charge on the ion.

Now, let's use this relationship to calculate the osmolarity of the solution.

Assume we have 1 L of solution.

We will set up a table for easy calculation.

$\boldsymbol{\underline{\text{Amt/meq"color(white)(mll)"Ion"color(white)(mll)ncolor(white)(mll)"Amt/mOsmol}}}$
$\textcolor{w h i t e}{m m l} 145 \textcolor{w h i t e}{m m m l} \text{Na"^"+} \textcolor{w h i t e}{m l l} 1 \textcolor{w h i t e}{m m m m m} 145$
$\textcolor{w h i t e}{m m m l l} 3 \textcolor{w h i t e}{m m m l} \text{K"^"+} \textcolor{w h i t e}{m m} 1 \textcolor{w h i t e}{m m m m m m l} 3$
$\textcolor{w h i t e}{m m m l l} 5 \textcolor{w h i t e}{m m m l} \text{Ca"^"2+} \textcolor{w h i t e}{m} 2 \textcolor{w h i t e}{m m m m m m l} 2.5$
$\underline{\textcolor{w h i t e}{m m l l} 153 \textcolor{w h i t e}{m m m l} \text{Cl"^"-} \textcolor{w h i t e}{m m} 1 \textcolor{w h i t e}{m m m m m} 153}$
$\textcolor{w h i t e}{m m m m m m m m m m m l} \text{TOTAL = 303.5}$

Note that there are 153 meq of cations and 153 meq of anions, as required.

$\text{Osmolarity" = "303.5 mOsmol"/"1 L" = "304 mOsmol/L}$

Since the solution is so dilute, its density is almost the same as that of water, and the mass of the water is approximately 1 kg.

∴ The concentration is also 304 mOsmol/kg.
The text passage provides a detailed explanation and calculation for determining the osmolality of a solution, including the conversion of equivalents to osmoles and the calculation of the osmolarity.
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Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
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# Question #daedf

Sep 5, 2017

The osmolality of the solution is 304 mOsmol/kg.

#### Explanation:

An osmole (Osmol) is 1 mol of particles that contribute to the osmotic pressure of a solution.

The problem here is how to convert equivalents to osmoles.

The point to remember is that 1 eq of anything reacts with or produces 1 eq of anything else

For example,

$\text{CaCl"_2 → "Ca"^"2+" + "2Cl"^"-}$
$\text{1 mol"color(white)(mm)"1 mol"color(white)(ml)"2 mol}$
$\text{2 eq"color(white)(mmll)"2 eq"color(white)(mml)"2 eq}$

We see that $\text{2 mol Cl"^"-" = "2 eq Cl"^"-}$, but $\text{1 mol Ca"^"2+" = "2 eq Ca"^"2+}$.

For ions, the rule is, $\text{1 mol" = ncolor(white)(l) "eq}$ or

$\boldsymbol{\text{1 eq" = 1/n color(white)(l)"mol}}$,

where $n$ is the absolute value of the charge on the ion.

Now, let's use this relationship to calculate the osmolarity of the solution.

Assume we have 1 L of solution.

We will set up a table for easy calculation.

$\boldsymbol{\underline{\text{Amt/meq"color(white)(mll)"Ion"color(white)(mll)ncolor(white)(mll)"Amt/mOsmol}}}$
$\textcolor{w h i t e}{m m l} 145 \textcolor{w h i t e}{m m m l} \text{Na"^"+} \textcolor{w h i t e}{m l l} 1 \textcolor{w h i t e}{m m m m m} 145$
$\textcolor{w h i t e}{m m m l l} 3 \textcolor{w h i t e}{m m m l} \text{K"^"+} \textcolor{w h i t e}{m m} 1 \textcolor{w h i t e}{m m m m m m l} 3$
$\textcolor{w h i t e}{m m m l l} 5 \textcolor{w h i t e}{m m m l} \text{Ca"^"2+} \textcolor{w h i t e}{m} 2 \textcolor{w h i t e}{m m m m m m l} 2.5$
$\underline{\textcolor{w h i t e}{m m l l} 153 \textcolor{w h i t e}{m m m l} \text{Cl"^"-} \textcolor{w h i t e}{m m} 1 \textcolor{w h i t e}{m m m m m} 153}$
$\textcolor{w h i t e}{m m m m m m m m m m m l} \text{TOTAL = 303.5}$

Note that there are 153 meq of cations and 153 meq of anions, as required.

$\text{Osmolarity" = "303.5 mOsmol"/"1 L" = "304 mOsmol/L}$

Since the solution is so dilute, its density is almost the same as that of water, and the mass of the water is approximately 1 kg.

∴ The concentration is also 304 mOsmol/kg.
user
The text passage provides a detailed explanation and calculation for determining the osmolality of a solution, including the conversion of equivalents to osmoles and the calculation of the osmolarity.

TYPE: SINGLE_QUESTION_WITH_ANSWER