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Esteban finds that a small bar of pure gold, with a density of 19.3 g/cm³, displaces 80 cm³ of water. He calculates that it's mass is 1,544 g. He then calculates that a smalller bar, with half the volume of the first bar, has a mass half a great. Is he correct? Explain your reasoning.

Respuesta :

Answer : Yes, he is correct.

Solution : Given,

Density of pure gold = [tex]19.3g/Cm^3[/tex]

The pure gold displaces [tex]80Cm^3[/tex] of water that means the Volume of pure gold is equal to the volume of water displaces.

Volume of pure gold = [tex]80Cm^3[/tex]

Now we have to calculate the Mass of pure gold.

Formula used :      [tex]Density=\frac{Mass}{Volume}[/tex]

By rearranging the formula, we get the mass of pure gold.

[tex]\text{ Mass of pure gold}=\text{ Density of pure gold}\times \text{ Volume of pure gold}=19.3g/Cm^3\times 80Cm^3=1544g[/tex]

So, The mass of pure gold is 1544 g.

Now we have to calculate the Mass of pure gold when the volume become half.

The density of pure gold remains same because density is an intensive property which means that no matter how much of the substance is present, its value remains same.

Volume of pure gold = [tex]40Cm^3[/tex]      

[tex]\text{ Mass of pure gold}=\text{ Density of pure gold}\times \text{ Volume of pure gold}=19.3g/Cm^3\times 40Cm^3=772g[/tex]

So, we conclude that when the volume become half then the mass also half.

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