The specific gravity of iron is $7.87$ and the density of water at $4.00^@"C"$ is $"1.00 g/cm"^3$. You can use this information to find the density of iron. Find the volume occupied by $"9.50 g"$ of iron?

Question

The specific gravity of iron is $7.87$ and the density of water at $4.00^@"C"$ is $"1.00 g/cm"^3$. You can use this information to find the density of iron. Find the volume occupied by $"9.50 g"$ of iron?

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Md Ariful Islam · Answer 1 · Apr 02, 2024 09:48:30

The idea here is that the specific gravity of a substance is calculated by dividing the of that substance by the density of a reference substance, which is usually water at $4^@"C"$.

$"SG" = rho_"substance"/rho_ ("H"_2"O at 4"^@"C")$

This means that in order to find the density of a substance, you must multiply its specific gravity by the density of the reference substance.

$rho_"Fe" = "SG" * rho_ ("H"_ 2"O at 4"^@"C")$

Since you are told that water at $4^@"C"$ has a density of $"1.00 g cm"^(-3)$ and that iron has a specific gravity of $7.87$, you can say that its density must be equal to

$rho_ "Fe" = 7.87 * "1/00 g cm"^(-3)$

$rho_ "Fe" = "7.87 g cm"^(-3)$

Now, the density of a substance tells you the mass of exactly $1$ unit of volume of that substance. In this case, the density of iron tells you that $:1 cm"^3$ of iron has a mass of $"7.87 g"$.

This means that a $"9.50-g"$ sample of iron will occupy a volume of

$9.50 color(red)(cancel(color(black)("g"))) * "1 cm"^3/(7.87color(red)(cancel(color(black)("g")))) = "1.21 cm"^3$

The answer is rounded to three .

The specific gravity of iron is $7.87$ and the density of water at $4.00^@"C"$ is $"1.00 g/cm"^3$. You can use this information to find the density of iron. Find the volume occupied by $"9.50 g"$ of iron?

1 Answers