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lesantik [10]
2 years ago
5

ethanol and benzene dissolve in each other. when ml of ethanol is dissolved in l of benzene, what is the mass of the mixture?

Chemistry
1 answer:
dsp732 years ago
8 0

The total mass of the mixture will be "958.5 g".

Calculation ,

According to the question, the density of both the mixtures will be:

Ethanol = 0.785 g/mL ( given )

Benzene = 0.880 g/mL ( given )

The volume of both the mixtures,

Ethanol = 100 mL ( given )

Benzene = 1.00 L or 1000 mL  ( given )

As we know,

density = mass/volume

now,

The mass of Ethanol will be:

mass = density × volume =  0.785 g/mL × 100 mL

mass  = 78.5 g

The mass of Benzene will be:

mass = density × volume =   0.880 g/mL  × 1000 mL

mass  = 880 g

hence,

The total mass of the mixture = 78.5 g + 880 g = 958.5 g

Missing data ,

density of Ethanol = 0.785 g/mL

density of Benzene = 0.880 g/mL

The volume of both the mixtures,

Ethanol = 100 mL

Benzene = 1.00 L

Learn more about mass of the mixture here:

brainly.com/question/15189339

#SPJ4

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Answer:

\large \boxed{\text{21.6 L}}

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:        180.16

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\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{24.5 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1360 mol C$_{6}$H$_{12}$O}_{6}

(b) Moles of CO₂

\text{Moles of CO}_{2} =\text{0.1360 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol CO}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.8159 mol CO}_{2}

(c) Volume of CO₂

We can use the Ideal Gas Law.

pV = nRT

Data:

p = 0.960 atm

n = 0.8159 mol

T = 37  °C

(i) Convert the temperature to kelvins

T = (37 + 273.15) K= 310.15 K

(ii) Calculate the volume

\begin{array}{rcl}pV &=& nRT\\\text{0.960 atm} \times V & = & \text{0.8159 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{310.15 K}\\0.960V & = & \text{20.77 L}\\V & = & \textbf{21.6 L} \\\end{array}\\\text{The volume of carbon dioxide is $\large \boxed{\textbf{21.6 L}}$}

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