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3241004551 [841]
3 years ago
6

A 0.150 M sodium chloride solution is referred to as a physiological saline solution because it has the same concentration of sa

lts as normal human blood. Calculate the mass of NaCl solute needed to prepare 275.0 mL of a physiological saline solution.
Chemistry
1 answer:
Anna71 [15]3 years ago
7 0
<h3>Answer:</h3>

2.41 g NaCl

<h3>Explanation:</h3>

We are given the;

Molarity of NaCl = 0.150 M

Volume of the solution = 275.0 mL

We are required to calculate the mass of NaCl solute;

<h3>Step 1: Number of moles of NaCl solute </h3>

Moles are calculated by;

Moles = Molarity × Volume

Therefore;

Number of moles of NaCl = 0.150 M × 0.275 L

                                           = 0.04125 moles

<h3>Step 2: Mass of NaCl solute </h3>

Mass of the compound is given by multiplying the number of moles by the molar mass of the compound.

Molar mass of NaCl = 58.44 g/mol

Therefore;

Mass of NaCl = 0.04125 moles × 58.44 g/mol

                       = 2.41 g

Thus, the mass of NaCl solute is 2.41 g

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0.4M is the concentration of a solution formed when 4.00 grams of sodium hydroxide is dissolved to make a 250cm cubed solution.

Molar concentration, also known as molarity, quantity concentration, or substance concentration. Molarity is a unit used to describe the amount of a substance in a solution expressed as a percentage of its volume. The number of moles per liter, denoted by the unit symbol mol/L or mol/dm³ in SI units, is the most often used unit denoting molarity. Because the volume of most solutions very minimally changes with temperature owing to thermal expansion, using molar concentration in thermodynamics is frequently not practical.

The formula for molarity,

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Explanation:

Hello.

In this case, according to the balanced chemical reaction, we can write the law of rate proportions:

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Thus, we proceed as follows:

a. Since the rate of oxygen production is 0.15 M/min, we can make the following setup:

\frac{r_{N_2O_5}}{-2}  =\frac{r_{O_2}}{1}\\\\r_{N_2O_5}=\frac{r_{O_2}}{-2}  =\frac{0.15M/min}{-2}\\\\ r_{N_2O_5}=-0.075M/min

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\frac{r_{N_2O_4}}{2}  =\frac{r_{O_2}}{1}\\\\r_{N_2O_4}=\frac{r_{O_2}}{2}  =\frac{0.15M/min}{2}\\\\ r_{N_2O_4}=0.075M/min

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