
The specific gravity of the stainless steel spherical balls used in ball-bearings is \[10.2\] . How many iron atoms are present in each ball of diameter 1 cm if the balls contain \[84\% \] iron by mass? The atomic mass of iron is 56.
A.\[4.12 \times {10^{21}}\]
B.\[4.82 \times {10^{22}}\]
C.\[3.82 \times {10^{22}}\]
D.None of the above.
Answer
557.4k+ views
Hint:First of all we need to calculate the density of steel using the formula for specific gravity. Then using density and volume of the sphere we will get the value of mass. The mass of iron balls will be the same as the mass of steel balls.
Complete step by step answer:
We know that the density of water is \[1{\text{ g}}/{\text{ml}}\] .
\[{\text{specific gravity}} = \dfrac{{{\text{Density of steel}}}}{{{\text{Density of water}}}}\]
Substituting the known values we will get:
\[{\text{10}}{\text{.2}} = \dfrac{{{\text{Density of steel}}}}{{1{\text{ g}}/{\text{ml}}}}\]
The density of steel will be \[10.2{\text{ g}}/{\text{mL}}\]
Now since this big iron steel ball has the diameter 1 cm that means radius will be half of the diameter that is \[0.5{\text{ cm}}\] .
The volume of spherical steel ball will be:
\[\dfrac{4}{3}{\text{\pi }}{{\text{r}}^3} = \dfrac{4}{3} \times 3.14 \times {\left( {0.5} \right)^3} = 0.52{\text{ c}}{{\text{m}}^3}\]
The density and the volume of this steel ball is known to us, hence the mass of the stell ball will be:
\[{\text{Mass}} = 0.52 \times 10.2 = 5.34{\text{ g}}\]
The number of moles of iron that can be accommodated in one steel ball can be calculated by dividing the mass of the ball to the molar mass of iron. The molar mass of iron is 56. Number of moles of iron will be:
\[{\text{No}}{\text{. of moles }} = \dfrac{{5.34}}{{56}} = 0.095\]
The ball contains \[84\% \] iron by mass and one mole contains \[6.022 \times {10^{23}}\] atoms of iron. Hence the number of moles of iron present will be:
\[\dfrac{{84}}{{100}} \times 0.095 \times 6.022 \times {10^{23}} = 4.82 \times {10^{22}}\]
Hence, option B is correct.
Note:
The specific gravity is defined as the ratio of density of the substance to the ration of the density of water. It is also known as the relative density as it is not absolute. It is calculated with respect to the density of water.
Complete step by step answer:
We know that the density of water is \[1{\text{ g}}/{\text{ml}}\] .
\[{\text{specific gravity}} = \dfrac{{{\text{Density of steel}}}}{{{\text{Density of water}}}}\]
Substituting the known values we will get:
\[{\text{10}}{\text{.2}} = \dfrac{{{\text{Density of steel}}}}{{1{\text{ g}}/{\text{ml}}}}\]
The density of steel will be \[10.2{\text{ g}}/{\text{mL}}\]
Now since this big iron steel ball has the diameter 1 cm that means radius will be half of the diameter that is \[0.5{\text{ cm}}\] .
The volume of spherical steel ball will be:
\[\dfrac{4}{3}{\text{\pi }}{{\text{r}}^3} = \dfrac{4}{3} \times 3.14 \times {\left( {0.5} \right)^3} = 0.52{\text{ c}}{{\text{m}}^3}\]
The density and the volume of this steel ball is known to us, hence the mass of the stell ball will be:
\[{\text{Mass}} = 0.52 \times 10.2 = 5.34{\text{ g}}\]
The number of moles of iron that can be accommodated in one steel ball can be calculated by dividing the mass of the ball to the molar mass of iron. The molar mass of iron is 56. Number of moles of iron will be:
\[{\text{No}}{\text{. of moles }} = \dfrac{{5.34}}{{56}} = 0.095\]
The ball contains \[84\% \] iron by mass and one mole contains \[6.022 \times {10^{23}}\] atoms of iron. Hence the number of moles of iron present will be:
\[\dfrac{{84}}{{100}} \times 0.095 \times 6.022 \times {10^{23}} = 4.82 \times {10^{22}}\]
Hence, option B is correct.
Note:
The specific gravity is defined as the ratio of density of the substance to the ration of the density of water. It is also known as the relative density as it is not absolute. It is calculated with respect to the density of water.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

