Question

Number of moles of water in $488g$ of $BaC{l_2} \cdot 2{H_2}O$ are ______
A. $2$
B. $4$
C. $3$
D. $5$

Answer 1

We have to first calculate the molar mass of $BaC{l_2} \cdot 2{H_2}O$. Then, we have to know the number of moles of water present in every mole of the compound. We have to calculate the moles of water for one gram of the compound. Then, using the given mass of the compound, moles of water present in one gram of the compound we could obtain the number of moles of water present in $488g$ of $BaC{l_2} \cdot 2{H_2}O$.

Complete step by step answer:
Given data contains,
Mass of $BaC{l_2} \cdot 2{H_2}O$ is $488g$.
We have to first determine the molar mass of $BaC{l_2} \cdot 2{H_2}O$.
The atomic mass of barium is $137.327g$.
The atomic mass of chlorine is $35.453g$.
The atomic mass of oxygen is $15.9994g$.
The atomic mass of hydrogen is $1.00794g$.
We can notice from the formula of barium chloride dihydrate that there is one atom of barium, two atoms of chlorine, four atoms of hydrogen and two atoms of oxygen.
So, the molar mass of $BaC{l_2} \cdot 2{H_2}O$ is calculated as,
Molar mass of $BaC{l_2} \cdot 2{H_2}O$=$137.327g + 2\left( {35.453g} \right) + 4\left( {1.00794g} \right) + 2\left( {15.9994g} \right)$
Molar mass of $BaC{l_2} \cdot 2{H_2}O$=$244.26g/mol$
We have calculated the molar mass of $BaC{l_2} \cdot 2{H_2}O$ is $244.26g/mol$.
Every mole of the given compound has two moles of water.
So, $244g$ of $BaC{l_2} \cdot 2{H_2}O$ has two moles of water.
So one gram of this compound could have $\dfrac{2}{{244}}$ moles of water.
So, $488g$ of this compound would contain $\dfrac{{\left( {2 \times 488} \right)}}{{244}}$ moles of water
$488g$ would contain four moles of water. Option (B) is correct.

Therefore, the option B is correct.

Note: An alternate way to calculate the number of moles of water in $BaC{l_2} \cdot 2{H_2}O$ is shown below,
Given data contains,
Mass of $BaC{l_2} \cdot 2{H_2}O$ is $488g$.
The molar mass of $BaC{l_2} \cdot 2{H_2}O$ is $244.26g/mol$.
The moles of $BaC{l_2} \cdot 2{H_2}O$ in $488g$=$\dfrac{{Given.weight}}{{molecular .weight}} = \dfrac{{488}}{{244}} = 2$
From the molecular formula, we know that $1mole$ of $BaC{l_2} \cdot 2{H_2}O$ contains two moles of water, so two moles of $BaC{l_2} \cdot 2{H_2}O$ would also have water=$2 \times 2 = 4$moles. Option (B) is correct.