Question

What is the mass of sodium in 42 g of NaCl?
(A) 12.74 g
(B) 16.51 g
(C) 25.58 g
(D) 33.02 g

Answer 1

As we know that the quantitative proportion or the major property of the entire matter can be referred to as mass. It is basically the bulk or the quantity of a matter and gives us quantitative information regarding any substance. So here we have to calculate the mass of sodium in 42 g of NaCl.
Formula used:
We will use the following formula:-
$n=\dfrac{w}{M}$
where,
n = number of moles
w = mass of the element or compound
M = molar mass of the compound or atomic mass of an element.

Complete answer:
Let us first calculate the number of moles of sodium chloride (NaCl) followed moles and mass of sodium as follows:-
-Calculation of moles of NaCl:-
Atomic mass of Na = 23 g/mol
Atomic mass of Cl = 35.5 g/mol
Molar mass of NaCl = (23 + 35.5) g/mol = 58.5 g/mol
The given value of mass of NaCl is (w) = 42 grams
On substituting all the values in the formula of number of moles, we get:-
$\begin{aligned} & \Rightarrow n=\dfrac{w}{M} \\\ & \Rightarrow n=\dfrac{42g}{58.5g/mol} \\\ & \Rightarrow n=0.717moles \\\ \end{aligned}$
-As we know that NaCl is composed of sodium and chlorine ion as shown below:-
$NaCl\to N{{a}^{+}}+C{{l}^{-}}$
This can be seen that 1 mole of NaCl is composed of 1 mole of sodium and 1 moles of chlorine element which means that 0.0717 moles of NaCl will be composed of 0.0717 moles of sodium and 0.0717 moles of chlorine element.
-Calculation of mass of Na in 42 grams of NaCl as follows:-
$n=\dfrac{w}{M}$
Let us rearrange the formula for further calculations:-
$\begin{aligned} & \Rightarrow w=M\times n \\\ & \Rightarrow w=23g/mol\times 0.717moles \\\ & \Rightarrow w=16.51grams \\\ \end{aligned}$
Therefore the mass sodium in 42 g of NaCl is (B) 16.51 g.

Note:
-Remember to put all the values in the formula along with their units so as to obtain better results with less number of errors and more accuracy which is required in physical chemistry.
-Also understand and practice the mole concepts and concentration related questions.