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.

Answer : Yes, he is correct.

Solution : Given,

Density of pure gold =

The pure gold displaces of water that means the Volume of pure gold is equal to the volume of water displaces.

Volume of pure gold =

Now we have to calculate the Mass of pure gold.

Formula used :      

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

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 =      

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

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

the volume increases