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sergey [27]
3 years ago
14

A 2.0% (m/v) nacl solution contains ________ g nacl in 300. ml of solution.

Chemistry
2 answers:
Sauron [17]3 years ago
8 0

when composition is given as m/v % , the value is the mass of the solute in 100 mL of the solution

this means there is 2.0 g in 100 mL of the solution

therefore if 100 mL of solution contains - 2.0 g

then 300 mL of the solution would contain - 2.0 g / 100 mL x 300 mL = 6.0 g

there's a mass of 6.0 g of NaCl in 300 mL of the soltuion

UNO [17]3 years ago
3 0

300 mL of NaCl solution contains \boxed{{\text{6 g}}} NaCl.

Further Explanation:

The concentration is the proportion of substance in the mixture. The most commonly used concentration terms are as follows:

1. Molarity (M)

2. Molality (m)

3. Mole fraction (X)

4. Parts per million (ppm)

5. Mass percent ((w/w)%)

6. Volume percent ((v/v)%)

7. Mass per volume percent ((w/v)%)

Mass per volume percent is defined as the amount of solute dissolved in a given volume of the solvent, multiplied by 100 %.

The formula to calculate the mass per volume percent of the solution is as follows:

{\text{Mass per volume}}\left({{\text{m/v}}}\right){\text{\%}}=\left({\frac{{{\text{Mass (g) of solute}}}}{{{\text{Volume (mL) of solution}}}}}\right)\times\left({100\;\%}\right)

Given information:

A 2.0 % (m/v) NaCl solution

To calculate:

Amount of NaCl in 300 mL of the solution

Firstly, we have to find out the amount of NaCl that is present in 2.0 % NaCl solution.

A 2.0 % NaCl solution indicates that 2 g of NaCl is dissolved in 100 mL of the solution.

So the amount of NaCl in 300 mL of the solution is calculated as follows:

\begin{gathered}{\text{Amount of NaCl}}=\left({\frac{{{\text{2 g}}}}{{{\text{100 mL}}}}}\right)\left({{\text{300 mL}}}\right)\\={\mathbf{6}}\;{\mathbf{g}}\\\end{gathered}

So 300 mL of NaCl solution contains 6 g of NaCl.

Learn more:

1. What is the mass of 1 mole of viruses: brainly.com/question/8353774

2. Determine the moles of water produced: brainly.com/question/1405182

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Concentration terms

Keywords: 2.0 %, NaCl, solution, concentration, proportion, substance, mixture, molarity, molality, mole fraction, parts per million, mass percent, volume percent, mass per volume percent, m, v, m/v, m/v %, 6 g.

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Answer:

Molecular formula => C₃H₈O₃

Explanation:

From the question given above, the following data were obtained:

Carbon (C) = 39.12%

Hydrogen (H) = 8.75%

Oxygen (O) = 51.12%

Molar mass of compound = 92.09 g/mol

Molecular formula =?

Next, we shall determine the empirical formula of the compound. This can be obtained as follow:

C = 39.12%

H = 8.75%

O = 51.12%

Divide by their molar mass

C = 39.12 / 12 = 3.26

H = 8.75 / 1 = 8.75

O = 51.12 / 16 = 3.195

Divide by the smallest

C = 3.26 / 3.195 = 1

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O = 3.195 / 3.195 = 1

Thus, the empirical formula is CH₂.₇O

Finally, we shall determine the molecular formula of the compound. This can be obtained as follow:

Empirical formula = CH₂.₇O

Molar mass of compound = 92.09 g/mol

Molecular formula =?

Molecular formula = Empirical formula × n

Molecular formula = [CH₂.₇O]ₙ

92.09 = [12 + (2.7×1) + 16] × n

92.09 = 30.7n

Divide both side by 30.7

n = 92.09 / 30.7

n = 3

Molecular formula = [CH₂.₇O]ₙ

Molecular formula = [CH₂.₇O]₃

Molecular formula = C₃H₈O₃

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