Question

Silver (atomic weight= $108g.mo{l^{ - 1}}$ ) has a density of $10.5g.c{m^{ - 3}}$ . The number of silver atoms on a surface of area ${10^{ - 12}}{m^2}$ can be expressed in scientific notation as $y \times {10^{ - x}}$. The value of $x$ is ?

Answer 1

To figure out how many atoms are there in a sample, we should first multiply the weight in grams by the (amu) atomic mass from the periodic table, then multiply by Avogadro's number: $6.02 \times {10^{23}}$.

Complete answer:
Firstly, let us look at the values that are given in the question:
Density= $10.5g.c{m^{ - 3}}$
Surface area= ${10^{ - 12}}{m^2} = {10^{ - 8}}c{m^2}$
$10.5gm$ of silver $\to$ $1c{m^3}$
Now, we need to convert the gram unit to mole. The formula used is:
$mole = \dfrac{{weight}}{{Molecular.weight}} \\\ mole = \dfrac{{10.5}}{{108}} \\\$

We know that,
Atom of silver $= density \times {N_A}$
Substituting the values,
Atom of silver= $\dfrac{{10.5}}{{108}} \times 6.023 \times {10^{23}}$
Now , in order to change the values from $cm$ to $m$, we will raise the power to $\dfrac{2}{3}$.
Hence,
Atom of silver= ${\left( {\dfrac{{10.5}}{{108}} \times 6.023 \times {{10}^{23}}} \right)^{\dfrac{2}{3}}}$
Now, the number of silver atom for $1c{m^2}$ surface area is:
${\left( {\dfrac{{10.5}}{{108}} \times 6.023 \times {{10}^{23}}} \right)^{\dfrac{2}{3}}} \to 1c{m^2} \\\ {\left( {\dfrac{{10.5}}{{108}} \times 6.023 \times {{10}^{23}}} \right)^{\dfrac{2}{3}}} \times {10^{ - 8}} \to {10^{ - 8}}c{m^2} \\\ = 1.5 \times {10^{ - 7}}atoms \\\$
Now the scientific notation is equal to the above value. Hence, we get:
$y \times {10^x} = 1.5 \times {10^{ - 7}}$
Comparing the values we get, $y = 1.5$ and $x = 7$.
Hence, the answer is $7$ .

Note:
The Avogadro constant is a proportionality element that connects the number of constituent particles in a sample to the volume of material present. The reciprocal mole is its SI unit, and it is defined as NA = $6.022 \times {10^{23}}mo{l^{ - 1}}$. It is named after Amedeo Avogadro, an Italian physicist.