Question

the mass of an atom of sodium is23u.calculate the number of sodium atoms in 1mg.
Take 1u=1.66×10 to the power -27kg.​

Answer 1

Given:

Supplied Mass of Sodium is 1 mg.

Atomic mass of Sodium is 23 u.

1 u = 1.66 × 10^(-27) kg.

To find:

Number of Sodium atoms in supplied mass.

Calculation:

Atomic mass of Sodium means that the mass of 1 atom of Sodium is 23 u.

$\sf{1 \: atom \: of \: sodium = 23u}$

$\sf{ = > 1 \: atom \: of \: sodium = 23\times 1.66 \times {10}^{ - 27} \: kg }$

$\sf{ = > 1 \: atom \: of \: sodium = 38.18\times {10}^{ - 27} \: kg }$

$\sf{ = > 1 \: atom \: of \: sodium = 38.18 \times {10}^{ - 27} \times {10}^{6} \: mg }$

$\sf{ = > 1 \: atom \: of \: sodium = 38.18\times {10}^{ - 21} \: mg }$

So, number of atoms in supplied mass:

$\rm{ \therefore \: no. \: of \: atoms = \dfrac{supplied \: mass}{mass \: of \: 1 \: atom} }$

$\rm{ = > \: no. \: of \: atoms = \dfrac{1}{38.18 \times {10}^{ - 21} } }$

$\rm{ = > \: no. \: of \: atoms = 2.6 \times {10}^{19} \: atoms }$

So , final answer is:

$\boxed{ \bold{\: no. \: of \: atoms = 2.6 \times {10}^{19} \: atoms }}$

Answer 2

Explanation:

Supplied Mass of Sodium is 1 mg.

Atomic mass of Sodium is 23 u.

1 u = 1.66 × 10^(-27) kg.

To find:

Number of Sodium atoms in supplied mass.

Calculation:

Atomic mass of Sodium means that the mass of 1 atom of Sodium is 23 u.

\sf{1 \: atom \: of \: sodium = 23u}1atomofsodium=23u

\sf{ = > 1 \: atom \: of \: sodium = 23\times 1.66 \times {10}^{ - 27} \: kg }=>1atomofsodium=23×1.66×10

−27

kg

\sf{ = > 1 \: atom \: of \: sodium = 38.18\times {10}^{ - 27} \: kg }=>1atomofsodium=38.18×10

−27

kg

\sf{ = > 1 \: atom \: of \: sodium = 38.18 \times {10}^{ - 27} \times {10}^{6} \: mg }=>1atomofsodium=38.18×10

−27

×10

6

mg

\sf{ = > 1 \: atom \: of \: sodium = 38.18\times {10}^{ - 21} \: mg }=>1atomofsodium=38.18×10

−21

mg

So, number of atoms in supplied mass:

\rm{ \therefore \: no. \: of \: atoms = \dfrac{supplied \: mass}{mass \: of \: 1 \: atom} }∴no.ofatoms=

massof1atom

suppliedmass

\rm{ = > \: no. \: of \: atoms = \dfrac{1}{38.18 \times {10}^{ - 21} } }=>no.ofatoms=

38.18×10

−21

1

\rm{ = > \: no. \: of \: atoms = 2.6 \times {10}^{19} \: atoms }=>no.ofatoms=2.6×10

19

atoms

So , final answer is:

\boxed{ \bold{\: no. \: of \: atoms = 2.6 \times {10}^{19} \: atoms }}

no.ofatoms=2.6×10

19

atoms