Question

$1.7g$ of $N{{a}_{2}}C{{O}_{3}}.n{{H}_{2}}O$ were dissolved in water and the volume was made up to $425ml$. $20$ml of this diluted solution required $20ml$$\dfrac{M}{20}$ HCl for complete neutralization. The value of n is
A. $7$
B. $3$
C. $2$
D. $10$

Answer 1

Hint: This question involves knowledge of various concepts. The concept of equivalent weight, number of milliequivalents, n-factor, molar mass and simple mathematical calculation. To solve this, we will firstly calculate the milliequivalents of given compound in $425$ml solution. From this information we will calculate the molar mass and finally will decipher the value of n.

Complete step-by-step answer:Let us firstly find the number of milliequivalents in $425$ ml of solution of given salt. But we know that no such information is mentioned which will help us to directly calculate the milliequivalents. So, we will firstly calculate the milliequivalents in $20$ml of given solution and then by simple unitary method will calculate the milliequivalents for 425ml.
We know that in neutralization the number of milliequivalents of acidic and basic solution is same
Therefore,
milliequivalents in $20$ml of $N{{a}_{2}}C{{O}_{3}}.n{{H}_{2}}O$= milliequivalents in $20\,ml\,\dfrac{M}{10}$HCl
We know that number of milliequivalents is given by formula
Number of milliequivalents = $Molarity\times Volume$
Therefore, number of milliequivalents of $20$ml HCl is calculated as
$\Rightarrow \dfrac{M}{20}\times 20=1$milliequivalents
Using unitary method, we can say that number of milliequivalents in $425ml$ solution is equal to
$\Rightarrow \dfrac{1}{20}\times 425=21.25$ milliequivalents
Now, we know that number of milliequivalents = $\dfrac{given mass}{equivalent\,mass}\times 1000$ …. equation 1
And we know that $equivalent mass=\,\dfrac{Molar\,mass}{n-factor}$ ……. Equation 2
Putting equation 2 in equation 1 we get,
Number of milliequivalents = $\dfrac{given\,mass\times n-factor\times 1000}{molar\,mass}$ …. Equation 3
Given mass= $1.7$grams
 number of milliequivalents = $21.25$
We know that $N{{a}_{2}}C{{O}_{3}}$will dissociate into $2N{{a}^{+}}\And C{{O}_{3}}^{2-}$. The ions have charge = $2$. And we know that in salt n-factor = charge on ions
Hence n-factor= $2$
Substituting the value of given mass, equivalents, n-factor in equation 3, we get
$21.25=\dfrac{1.7\times 2\times 1000}{molar\,mass}$
Therefore, molar mass= $\dfrac{3400}{21.25}=160$g/mole
This is the molar mass of given compound sodium carbonate.
Also, we know that molar mass of $N{{a}_{2}}C{{O}_{3}}. n{{H}_{2}}O$ can be written as = $106+18n$
Comparing both the above equation, we get
$\begin{align}
  & \Rightarrow 106+18n=160 \\
 & \Rightarrow 18n=160-106 \\
 & \Rightarrow 18n=54 \\
 & \Rightarrow n=\dfrac{54}{18}=3 \\
\end{align}$
Therefore, the value of n=$3$

Hence, the correct option is B, 3

Note:The most important point for the question was interrelation of concept. In physical chemistry questions always see what is given and then determine the starting and ending point. Then try to interrelate the concepts to reach a solution. This interrelation of concepts comes with practice.