# Which sample contains the least number of atoms: "2 g Ar; 2 g C; 2 L He; 2 L dihydrogen gas"?

Jul 31, 2017

Which question does you want answered?

#### Explanation:

Clearly the last option is the most suitable, and most safe:

$N a O H \left(a q\right) + H N {O}_{3} \left(a q\right) \rightarrow N {a}^{+} N {O}_{3}^{-} \left(a q\right) + {H}_{2} O \left(l\right)$

The resultant sodium nitrate could be crystallized from solution.

The other options involve addition of sodium to water, or acid, which would not be the safest exercise.

If you want to know which of the following contains LEAST amount of atoms, i.e. out of $\text{2 g Ar}$; $\text{2 g C}$; $2 \cdot L$ $\text{He}$, and $2 \cdot L$ ${H}_{2}$, we must assume standard conditions of temperature and pressure to assess the gaseous quantities, and we know that the molar volume of a gas is approx. $25 \cdot L \equiv 25 \cdot {\mathrm{dm}}^{3}$ given the assumption.

$\text{Number of argon atoms}$ $=$ $\frac{2.0 \cdot g}{40.0 \cdot g \cdot m o {l}^{-} 1} \times {N}_{A} = 5 \times {10}^{-} 2 \cdot {N}_{A}$.

$\text{Number of carbon atoms}$ $=$ $\frac{2.0 \cdot g}{12.01 \cdot g \cdot m o {l}^{-} 1} \times {N}_{A} = 1.67 \times {10}^{-} 1 \cdot {N}_{A}$.

$\text{Number of helium atoms}$ $=$ $\frac{2.0 \cdot L}{25.4 \cdot L \cdot m o {l}^{-} 1} \times {N}_{A} = 7.87 \times {10}^{-} 2 \cdot {N}_{A}$.

$\text{Number of hydrogen atoms}$ $=$ $\frac{2 \times 2.0 \cdot L}{25.4 \cdot L \cdot m o {l}^{-} 1} \times {N}_{A} = 1.57 \times {10}^{-} 1 \cdot {N}_{A}$.

In each case, ${N}_{A} \equiv \text{the Avocado Number} = 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$.

The least number of atoms is expressed in the sample of argon gas. Do you agree?