Question

When $0.575\times {{10}^{-2}}$ Kg of glauber's salt ($N{{a}_{2}}S{{O}_{4}}.10{{H}_{2}}O$) is dissolved in 3g water, we get 1$d{{m}^{3}}$ of a solution of density 1077.2Kg${{m}^{-3}}$. Calculate the molarity, molality and mole fraction in the solution.
(A) Molarity(M)= 0.2502, Molality(m) = 0.2402, Mole fraction = \[4.304~\times {{10}^{-3}}\]
(B) Molarity(M)= 0.2402, Molality(m) = 0.2502, Mole fraction = \[4.304~\times {{10}^{-3}}\]
(C) Molarity(M)= 0.2502, Molality(m) = 0.2402, Mole fraction = \[4.304~\times {{10}^{-1}}\]
(D)None

Answer 1

Hint: To solve this question we should know the formula to calculate molarity, molality and mole fraction. A very important thing to observe in question is that we have to calculate above mentioned quantities of $NaS{{O}_{4}}$ in glauber's salt ($N{{a}_{2}}S{{O}_{4}}.10{{H}_{2}}O$).

Complete step by step solution:
Firstly let's know the definition and formula to calculate molarity, molality and mole fraction.
The Glauber's salt is $N{{a}_{2}}S{{O}_{4}}.10{{H}_{2}}O$.
Given mass of Glauber's salt = $0.575\times {{10}^{-2}}$Kg = 5.75g
In 1$d{{m}^{3}}$of solution, the weight of Glauber's salt $N{{a}_{2}}S{{O}_{4}}.10{{H}_{2}}O$is 5.75g.
Volume = $\dfrac{mass}{Density}$ ………..equation 1
From equation 1 we can calculate the mass of solution
Therefore, the mass of the solution = volume $\times $density
$\implies$ 1$d{{m}^{3}}$ $\times $ 1077.2$\times $ ${{10}^{-3}}$$Kgd{{m}^{-3}}$
$\implies$ 1.0772$\times $${{10}^{-3}}$kg
$\implies$ 1.0772g
Let's find the mass of $NaS{{O}_{4}}$ in 5.75g of $N{{a}_{2}}S{{O}_{4}}.10{{H}_{2}}O$
Molecular weight of $N{{a}_{2}}S{{O}_{4}}.10{{H}_{2}}O$= 322g
Molecular weight of $NaS{{O}_{4}}$= 142g
From this, we can say that 322g of Glauber's salt ($N{{a}_{2}}S{{O}_{4}}.10{{H}_{2}}O$) contains 142g of $NaS{{O}_{4}}$
So, $NaS{{O}_{4}}$ in 5.75g of $N{{a}_{2}}S{{O}_{4}}.10{{H}_{2}}O$is,
$\implies$ \[\dfrac{142\times 5.75}{322}\]g
$\implies$ 2.5371g
Number of moles of $NaS{{O}_{4}}$= $\dfrac{Weight\text{ }of\text{ }NaS{{O}_{4}}\text{ }in\text{ }grams}{Molecular\text{ }weight}$
$\implies$ $\dfrac{2.5371}{142}$
$\implies$ 0.0178
Thus, mole fraction of $NaS{{O}_{4}}$is 0.0178
Molarity is also known as molar concentration which is defined as the number of moles dissolved in 1litre.
Molarity = $\dfrac{Number\text{ }of\text{ }moles\text{ }of\text{ }NaS{{O}_{4}}(Solute)}{Volume\text{ }of\text{ }solution\text{ }in\text{ }litre}$
$\implies$ $\dfrac{0.0178}{0.001\times {{10}^{3}}}$
$\implies$ 0.0178M
Molality can be defined as the number of moles of solute present in kg (1000g)
Molality = $\dfrac{Molarity}{kg}$
$\implies$ \[\dfrac{2.5371}{142}\times \dfrac{1}{3}\times 1000\]
$\implies$ 5.955m

Hence option D is the correct answer.

Note: It is very important to understand the difference between molarity and molality. Molarity is the total number of moles of solute per liter and molality is the total number of moles of solute per kilogram. Always mention units because not mentioning units may lead to wrong answers.