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Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

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BuyFindarrow_forward

Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
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89. A mixture is prepared by mixing 50.0 g of ethanol, 50.0 g of water, and 5.0 g of sugar. What is the mass percent of each component in the mixture? How many grams of the mixture should one take in order to have 1 .5 g of sugar? How many grams of the mixture should one take to have 10.0 g of ethanol?

Interpretation Introduction

Interpretation:

The mass percentage of each component of the given mixture is to be calculated. The mass of the given mixture required to have 1.5g of sugar is to be calculated. The mass of the given mixture required to have 10.0g of ethanol is to be calculated.

Concept Introduction:

The mass percentage of a component present in a mixture is calculated by dividing the mass of the component present in the mixture by the total mass of the mixture and then multiplying it by 100. The formula for mass percentage is represented as:

mass%=MMt×100%

Where,

  • M represents the mass of the component present in the mixture.
  • Mt represents the total mass of mixture.
Explanation

The mass of ethanol in the mixture is 50.0g.

The mass of water in the mixture is 50.0g.

The mass of sugar in the mixture is 5.0g.

The formula for mass percentage is represented as:

mass%=MMt×100%    (1)

Where,

  • M represents the mass of the component present in the mixture.
  • Mt represents the total mass of mixture.

The total mass of the mixture is given as:

Mt=iMi

Where,

  • Mi represents the mass of the ith component of the mixture.

Substitute the value of mass of ethanol, water and sugar in the above equation.

Mt=50.0g+50.0g+5.0g=105.0g

The total mass of the mixture is 105.0g.

Substitute the value of mass of ethanol and total mass of the mixture in the equation (1).

mass%=50.0g100%105.0g=47.62%

The mass percentage of ethanol in the mixture is 47.62%.

Substitute the value of mass of water and total mass of the mixture in the equation (1).

mass%=50.0g100%105.0g=47.62%

The mass percentage of water in the mixture is 47

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