Question

The density of gold is $19g/c{{m}^{3}}$. If $1.9\times {{10}^{-4}}g$ of gold is dispersed in one litre of water to give a sol having spherical gold particles of radius $10nm$, then the number of gold particles per $m{{m}^{3}}$ of the sol will be:
A. $1.9\times {{10}^{12}}$
B. $6.3\times {{10}^{14}}$
C. $6.3\times {{10}^{10}}$
D. $2.4\times {{10}^{6}}$

Answer 1

Hint: Think about how we can relate the given masses and volumes of the gold while disregarding the use of the terminology sol since it is extra information and will not actually help us in solving the problem.

Complete answer:
We will be using the basic formula that relates volume, density, and mass to each other. If we calculate the total volume of gold given to us and then calculate the volume of each spherical particle, we will be able to find the total number of gold particles that are suspended in the sol. After finding the total number, we can easily find the number of particles present per unit volume since the total volume of the sol has been given to us.
The formula that we are going to use is:
\[\begin{align}
  & density=\dfrac{mass}{volume} \\
 & volume=\dfrac{mass}{density} \\
\end{align}\]
We have been given the density and the mass of the gold that is dispersed in the sol. From this, we will calculate the total volume of the gold. The given density is $19g/c{{m}^{3}}$ and the given mass is $1.9\times {{10}^{-4}}g$. Now, putting these in the formula, we get the total volume of gold as:
\[\begin{align}
  & volume=\dfrac{1.9\times {{10}^{-4}}g}{19g/c{{m}^{3}}} \\
 & volume={{10}^{-5}}c{{m}^{3}} \\
\end{align}\]
Now, we will find the volume of each spherical gold particle that will be present in the sol. The radius of each particle is given as $10nm$ which is equal to ${{10}^{-6}}cm$. We know that the volume of a sphere is given by:
\[volume=\dfrac{4}{3}\pi {{r}^{3}}\]
Now, putting the value of $r$ in the formula, we get:
\[\begin{align}
  & volume=\dfrac{4\times 3.142\times {{({{10}^{-6}}cm)}^{3}}}{3} \\
 & volume=4.189\times {{10}^{-18}}c{{m}^{3}} \\
\end{align}\]
Now we can divide the total volume of the gold by the volume of one spherical particle to get the total number of particles in the sol.
\[\begin{align}
  & \text{Total gold particles in the sol =}\dfrac{{{10}^{-5}}}{4.189\times {{10}^{-18}}} \\
 & \text{Total gold particles in the sol}=0.238\times {{10}^{13}} \\
 & \text{Total gold particles in the sol}=2.38\times {{10}^{12}} \\
\end{align}\]
So, there are $2.38\times {{10}^{12}}$ particles present in $1000c{{m}^{3}}$. According to this, we will calculate the number of particles present in $1m{{m}^{3}}$ of the sol.
The number of particles present in $1c{{m}^{3}}$ of the sol is $2.38\times {{10}^{9}}$. And the number of particles present in $1m{{m}^{3}}$ of the sol will be $2.38\times {{10}^{6}}$.

Hence, the correct answer to this question is ‘D. $2.4\times {{10}^{6}}$’.

Note:
Recall that 1 liter is equal to ${{10}^{-3}}{{m}^{3}}$ of volume. This value can be further converted into other units like $m{{m}^{3}}$. Take into consideration that although while converting $cm$ to $mm$ we divide by a factor of only 10. But here, we are considering volume, so the value should be divided by ${{10}^{3}}$ to get the corresponding value in $m{{m}^{3}}$.