Question

What is the mass of phosphorus that contains twice the number of atoms found in 14 g of iron? [Relative atomic mass : P = 31; Fe = 56)

А 62.0 g

B 28.0 g

C 15.5 g

D 10.7 g​

Answer 1

Answer:

The mass of phosphorus that contains twice the number of atoms found in $\[14g\]$ of iron is $\[{\rm{15}}{\rm{.5g of P }}\]$.

Explanation:

Atomic number of iron is 56 and that of phosphorus is 31.

$\[{\rm{1mole Fe = 56g}}\]$

$\[{\rm{1 mole P = 31g}}\]$

One mole of any element contains $\[{6.022 \times {{10}^{23}}}\]$ atoms.

If $\[56g\]$ of iron contains $\[{6.022 \times {{10}^{23}}}\]$ atoms, then $\[{\rm{14g of Fe}}\]$ will contain:

$\[{\rm{14g of Fe = }}\frac{{6.022 \times {{10}^{23}}}}{{56}} \times 14 = 1.5 \times {10^{23}}atoms\]$

The number of atoms present in $\[{\rm{62g of P}}\]$ is:

$\[{\rm{62g of P = }}\frac{{6.022 \times {{10}^{23}}}}{{31}} \times 62 = 1.2 \times {10^{24}}atoms\]$

The number of atoms present in $\[{\rm{28g of P}}\]$ is:

$\[{\rm{28g of P = }}\frac{{6.022 \times {{10}^{23}}}}{{31}} \times 28 = 5.4 \times {10^{23}}atoms\]$

The number of atoms present in $\[{\rm{15}}{\rm{.5g of P }}\]$ is:

$\[{\rm{15}}{\rm{.5g of P = }}\frac{{6.022 \times {{10}^{23}}}}{{31}} \times 15.5 = 3.0 \times {10^{23}}atoms\]$

The number of atoms present in $\[{\rm{10}}{\rm{.7g of P}}\]$ is:

$\[{\rm{10}}{\rm{.7g of P = }}\frac{{6.022 \times {{10}^{23}}}}{{31}} \times 10.7 = 2.07 \times {10^{23}}atoms\]$

Therefore, the mass of phosphorus that contains the twice the number of atoms found in $\[{\rm{14g of Fe }}\]$ is $\[{\rm{15}}{\rm{.5g of P }}\]$.