Question

The lightest metal known is
(A) beryllium
(B) lithium
(C) sodium
(D) boron

Answer 1

Hint: A light metal is any metal of relatively low density.
 density = ${\displaystyle \frac{mass}{volume}}$
As we can see density is directly proportional to mass (in this case, atomic mass). So, we can compare different metals on the basis of atomic mass.
Less will be the atomic mass, lee will be density, lighter will be the metal.

Complete solution step by step:
Atomic mass of beryllium = $9$ u
Atomic mass of lithium = $7$ u
Atomic mass of sodium = $23$ u
Atomic mass of boron = $11$ u
We can clearly observe here that the atomic mass of lithium is least among these ( $7$ u). Therefore, it will be having the least density and so will be the lightest metal.
So, the correct option would be B: lithium.

Additional information:
When we talk about a material being lightweight, what we are really discussing is density. Density is the amount of matter per unit of volume. The less matter is there in space, the lighter the substance is. So, the lightest metallic element is the one with the lowest density. The lightest or least dense element which is a metal is lithium. The atomic number of lithium is $3$ u. Among all the alkali metals, lithium is highly reactive because of its small size. The reason why alkali metals are so light is because these elements are the first atoms in their period on the periodic table to gain an additional electron shell.
In a periodic table, on moving down the group, the atomic radius increases resulting in an increase in density of elements. On moving from left to right in a period, atomic radius decreases resulting in decrease in the density of the elements.

Note:
Lithium is the lightest metal because it is placed in the first group of periodic tables (in alkali metal group) and it has the lowest atomic weight. We can also solve this problem by the concept of atomic number. Less will be atomic number, lighter the element will be. However, there are some exceptions too which should be kept in mind.