Question

One uranium atom has a mass of $3.95\, \times {10^{ - 22}}$ grams. How would you work out an estimate for the number of uranium atoms in \[1\,Kg\]of uranium?

Answer 1

Hint: Here, we will use the concept of Avogadro number and molar mass. We know that Avogadro number is equal to $6.022\, \times {10^{23}}$ particles. So, estimating that if one mole that is mass of $6.022\, \times {10^{23}}$ atoms is equals to given mass $3.95\, \times {10^{ - 22}}$ grams then, we can easily calculate the number of atoms present in \[1\,Kg\] of uranium.

Complete step-by-step answer:
For solving these types of questions in which we have mass and atoms types of word problems. In these types of questions, the main thing which you have to keep in mind is the relation. A relation that one atom if has a certain amount of mass then for one mole of it requires the multiplication of mass with the number of atoms. Let’s start solving the question by taking one uranium mass.

We know that one mole is equal to $6.022\, \times {10^{23}}$ atoms so, writing it as$One\,mole\, = \,6.022 \times {10^{23}}\,atoms\,$
And it is given in the question that one atom has mass of $3.95\, \times {10^{ - 22}}$ grams$(3.95\, \times {10^{ - 22}})\,grams = \,Mass\,of\,1\,atom$
For ${N_A}$ particles we have to multiply the mass with $6.022\, \times {10^{23}}$ amount so,
$\,(3.95\, \times {10^{ - 22}} \times 6.022\, \times {10^{23}})\,grams\,consists\, = \,\,6.022\, \times {10^{23}}\,atoms$
$238\,grams\,consists\, = \,6.022\, \times {10^{23}}\,atoms$
Now in the question it was asking us for the amount of atoms that we will have in one kilograms. We know that one kilogram is equal to $1000\,g$ so let’s try to calculate $1000\,g$ .
$\Rightarrow 1\,grams\,consists\, = \,\left( {\dfrac{{6.022\, \times {{10}^{23}}\,}}{{238\,}}} \right)atoms$
$\Rightarrow 1000\,grams\,consists\, = \,\left( {\dfrac{{6.022\, \times {{10}^{23}}\,}}{{238\,}}} \right) \times 1000\,atoms$
$\Rightarrow 2.53\, \times {10^{24}}\,atoms$
From the above calculations we get the exact number of atoms which will present in one kilogram of uranium.

Note: Don’t forgot to multiply the given mass $3.95\, \times {10^{ - 22}}$ grams by Avogadro number $6.022\, \times {10^{23}}$ atoms because the mass of one mole is $3.95\, \times {10^{ - 22}}$ and $6.022\, \times {10^{23}}$ atoms consists one mole. Any element for which we are taking one mole of it, means it has ${N_A}$ particles and these particles form the whole mass.