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Question: On heating \(5\times {{10}^{-3}}\) equivalent of \(CaC{{l}_{2}}.x{{H}_{2}}O\), 0.18 gm of water is o...

On heating 5×10−35\times {{10}^{-3}} equivalent of CaCl2.xH2OCaC{{l}_{2}}.x{{H}_{2}}O, 0.18 gm of water is obtained. What is the value of x?

Explanation

Solution

The number of water molecules in the hydrated molecule can be determined from its equivalent weight, which encompasses the molar mass of the molecule and the n-factor.

Complete step by step answer:
Given the number of grams equivalent of hydrated calcium chloride compound, which on heating loses its water molecules to form anhydrous calcium chloride. It describes the number of electrons or ions transferred during the reaction.
It is obtained from the formula as follows:
No. of gram equivalents=Given weightequivalent weight of the molecule\text{No}\text{.}\,\text{of}\,\text{gram}\,\text{equivalents=}\dfrac{\text{Given}\,\text{weight}}{\text{equivalent weight of the molecule}} --------- (a)
where the equivalent weight of the hydrated calcium chloride will be molarmassn−factor=40+(2×35.5)+x(2×1+16)2=55.5+9x\dfrac{molar mass}{n-factor}=\dfrac{40+(2\times 35.5)+x(2\times 1+16)}{2}=55.5+9x
Then, substituting the value of equivalent weight and the given mass in equation (a), we will obtain the value of x, as follows:
5×10−3=0.1855.5+9x5\times {{10}^{-3}}=\dfrac{0.18}{55.5+9x}
55.5+9x=0.18×25×10−3=7255.5+9x=\dfrac{0.18\times 2}{5\times {{10}^{-3}}}=72
Then, x=72−55.59=1.8≈2x=\dfrac{72-55.5}{9}=1.8\approx 2

Therefore, the hydrated form has two water molecules in it, such that the hydrated form is CaCl2.2H2OCaC{{l}_{2}}.2{{H}_{2}}O.

Additional information: The calcium chloride is hygroscopic in nature, that is its tendency to hold or absorb the water molecule. Thus, it is present in various hydrated forms with 0, 1, 2, 4, and 6 molecules of water in the compound.
The dihydrate form is highly soluble in water and is a solid white crystalline ionic halide at room temperature.

Note: The number of gram equivalents is used in the determination of the concentration of the compound in per litre of solution in terms of normality. The dihydrate form is used in managing the dust and ice on roads and also for the refrigeration of the plants.