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goldfiish [28.3K]
3 years ago
12

A solution is prepared by mixing 631 mLmL of methanol with 501 mLmL of water. The molarity of methanol in the resulting solution

is 14.29 MM. The density of methanol at this temperature is 0.792 g/mLg/mL. Calculate the difference in volume between this solution and the total volume of water and methanol that were mixed to prepare the solution
Chemistry
2 answers:
Brrunno [24]3 years ago
7 0

<u>Answer:</u> The difference between the volume of solution and mixture of methanol and water is 0.039 L

<u>Explanation:</u>

We are given:

Volume of methanol = 631 mL

Volume of water = 501 mL

Volume of methanol + water = [631 + 501] mL = 1132 mL = 1.132 L   (Conversion factor:  1 L = 1000 mL)

To calculate the mass of methanol, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of methanol = 0.792 g/mL

Volume of methanol = 631 mL

Putting values in above equation, we get:

0.792g/mL=\frac{\text{Mass of methanol}}{631mL}\\\\\text{Mass of methanol}=(0.792g/mL\times 631mL)=499.75g

  • To calculate the volume of solution, we use the equation used to calculate the molarity of solution:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

Given mass of methanol = 499.75 g

Molar mass of methanol = 32 g/mol

Molarity of solution = 14.29 M

Putting values in above equation, we get:

14.29mol/L=\frac{499.75g}{32g/mol\times V}\\\\V=\frac{499.75}{32\times 14.29}=1.093L

Calculating the difference between the volumes:

\Delta V=V_{mixture}+V_{solution}

Putting values in above equation, we get:

\Delta V=[1.132-1.093]L=0.039L

Hence, the difference between the volume of solution and mixture of methanol and water is 0.039 L

just olya [345]3 years ago
5 0

Answer:

The difference in volume is 40 mL or 0.040 L

Explanation:

Step 1: Data given

Volume of methanol = 631 mL = 0.631 L

Volume of water = 501 mL = 0.501 L

Molarity of methanol = 14.29 M

Density of methanol = 0.792 g/mL

Step 2: Calculate mass methanol

Mass = density * volume

Mass methanol = 0.792 g/mL * 631 mL methanol = 499.8 g methanol

Step 3: Calculate moles methanol

Moles = mass / molar mass

Moles methanol = 499.8 g methanol / 32.04 g/mol

Moles methanol = 15.6 moles methanol

Step 4: Calculate volume

We should have 15.6 moles of methanol, and we need 14.29 M solution

Volume = moles / molarity

Volume =  15.6 moles / 14.29 M

Volume = 1.092 L

Step 5: Calculate the difference in volume

So we should have 1.092 Liters of solution, but we haf 1.132 Liters of solution:

1.132 L - 1.092 L =  0.040 L or 40 mL

The difference in volume is 40 mL or 0.040 L

So the difference is 39 mL.

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4 0
3 years ago
A 3.50 g sample of an unknown compound containing only C , H , and O combusts in an oxygen‑rich environment. When the products h
statuscvo [17]

Explanation:

First, calculate the moles of CO_{2} using ideal gas equation as follows.

                PV = nRT

or,          n = \frac{PV}{RT}

                = \frac{1 atm \times 4.41 ml}{0.0821 Latm/mol K \times 293 K}      (as 1 bar = 1 atm (approx))

                = 0.183 mol

As,   Density = \frac{mass}{volume}

Hence, mass of water will be as follows.

                Density = \frac{mass}{volume}

             0.998 g/ml = \frac{mass}{3.26 ml}    

                 mass = 3.25 g

Similarly, calculate the moles of water as follows.

        No. of moles = \frac{mass}{\text{molar mass}}

                              =  \frac{3.25 g}{18.02 g/mol}            

                              = 0.180 mol

Moles of hydrogen = 0.180 \times 2 = 0.36 mol

Now, mass of carbon will be as follows.

       No. of moles = \frac{mass}{\text{molar mass}}

          0.183 mol =  \frac{mass}{12 g/mol}            

                              = 2.19 g

Therefore, mass of oxygen will be as follows.

              Mass of O = mass of sample - (mass of C + mass of H)

                                = 3.50 g - (2.19 g + 0.36 g)

                                = 0.95 g

Therefore, moles of oxygen will be as follows.

          No. of moles = \frac{mass}{\text{molar mass}}

                               =  \frac{0.95 g}{16 g/mol}            

                              = 0.059 mol

Now, diving number of moles of each element of the compound by smallest no. of moles as follows.

                         C              H           O

No. of moles:  0.183        0.36       0.059

On dividing:      3.1           6.1            1

Therefore, empirical formula of the given compound is C_{3}H_{6}O.

Thus, we can conclude that empirical formula of the given compound is C_{3}H_{6}O.            

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The n and l values and the number of orbitals for sublevel 5g is :

5g shell , n= 5 subshell g , l = 4, Number of orbitals for sublevel = 9.

There are total four quantum numbers:

1) Principal quantum number , n

2) Angular quantum number , l

3) Magnetic quantum number , ml

4) spin quantum number , ms

For 5g shell, n = 5

subshell g , l = 4     ....0 - s , 1 - p , 2 - d, 3 - f, 4 -g

number of orbitals in subshell = (2l + 1)  ( 2×4 + 1) = 9

Thus,  The n and l values and the number of orbitals for sublevel 5g is :

5g shell , n= 5 subshell g , l = 4, Number of orbitals for sublevel = 9.

To learn more about quantum numbers here

brainly.com/question/14650894

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6 0
1 year ago
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vivado [14]

Answer:

The molar concentration of HCl in the aqueous solution is 0.0131 mol/dm3

Explanation:

To get the molar concentration of a solution we will use the formula:

<em>Molar concentration = mass of HCl/ molar mass of HCl</em>

<em></em>

Mass of HCl in the aqueous solution will be 40% of the total mass of the solution.

We can extract the mass of the solution from its density which is 1.2g/mL

We will further perform our analysis by considering only 1 ml of this aqueous solution.

The mass of the substance present in this solution is 1.2g.

<em>The mass of HCl Present is 40% of 1.2 = 0.48 g.</em>

The molar mass of HCl can be obtained from standard tables or by adding the masses of Hydrogen (1 g) and Chlorine (35.46 g) = 36.46g/mol

Therefore, the molar concentration of HCl in the aqueous solution is 0.48/36.46 = 0.0131 mol/dm3

7 0
3 years ago
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