Answer
Verified
391.5k+ views
Hint: Remember that the hydrate form of any compound A is given by $A.{{\operatorname{xH}}_{2}}O$ whereas the anhydrous forms of compound do not involve any water molecules in the chemical formula. The number of water molecules involved in the hydrated form depends from compound to compound.
Complete step by step solution:
Let us first observe the physical and chemical properties of Copper Sulphate before moving onto the specifics of this question.
Copper (II) sulphate, also known as copper sulphate, is an inorganic compound with the chemical formula \[CuS{{O}_{4}}.\operatorname{x}({{H}_{2}}O)\], where x can range from 0 to 5. The pentahydrate ($x=5$) is the most common form. Older names for this compound include blue vitriol, bluestone, vitriol of copper, and Roman vitriol.
The pentahydrate (\[CuS{{O}_{4}}\cdot 5{{H}_{2}}O\]), is the most commonly encountered salt and is bright blue in colour. It exothermically dissolves in water to give the aqua complex \[{{[Cu{{({{H}_{2}}O)}_{6}}]}^{2+}}\], which has an octahedral molecular geometry. The structure of the solid pentahydrate reveals a polymeric structure wherein copper is again octahedral but bound to four water ligands.
Anhydrous copper sulphate on the other hand is a grey powder. It is commonly used as a test for water. If it turns blue (which is chemically its hydrated form) when subjected to any liquid, then that liquid is water.
The interconversion of the hydrated and unhydrated form is fairly simple. When water is added to the unhydrated form (greyish white) it changes into its hydrated form (blue). To change back you just have to heat it again. The reaction is as follows:
\[CuS{{O}_{4}}\cdot 5{{H}_{2}}OCuS{{O}_{4}}+5{{H}_{2}}O\]
Note:
The hydrated form of copper sulphate should not be interpreted as its chemical structure. The water is added in the crystal lattice in between the molecules of copper sulphate. The form\[CuS{{O}_{4}}\cdot 5{{H}_{2}}O\] is actually a double salt. It can be proved by dissolving this in water. You can experimentally determine that once it is dissolved, the water molecules present in the crystal lattice are gone into the solution.
Complete step by step solution:
Let us first observe the physical and chemical properties of Copper Sulphate before moving onto the specifics of this question.
Copper (II) sulphate, also known as copper sulphate, is an inorganic compound with the chemical formula \[CuS{{O}_{4}}.\operatorname{x}({{H}_{2}}O)\], where x can range from 0 to 5. The pentahydrate ($x=5$) is the most common form. Older names for this compound include blue vitriol, bluestone, vitriol of copper, and Roman vitriol.
The pentahydrate (\[CuS{{O}_{4}}\cdot 5{{H}_{2}}O\]), is the most commonly encountered salt and is bright blue in colour. It exothermically dissolves in water to give the aqua complex \[{{[Cu{{({{H}_{2}}O)}_{6}}]}^{2+}}\], which has an octahedral molecular geometry. The structure of the solid pentahydrate reveals a polymeric structure wherein copper is again octahedral but bound to four water ligands.
Anhydrous copper sulphate on the other hand is a grey powder. It is commonly used as a test for water. If it turns blue (which is chemically its hydrated form) when subjected to any liquid, then that liquid is water.
The interconversion of the hydrated and unhydrated form is fairly simple. When water is added to the unhydrated form (greyish white) it changes into its hydrated form (blue). To change back you just have to heat it again. The reaction is as follows:
\[CuS{{O}_{4}}\cdot 5{{H}_{2}}OCuS{{O}_{4}}+5{{H}_{2}}O\]
Note:
The hydrated form of copper sulphate should not be interpreted as its chemical structure. The water is added in the crystal lattice in between the molecules of copper sulphate. The form\[CuS{{O}_{4}}\cdot 5{{H}_{2}}O\] is actually a double salt. It can be proved by dissolving this in water. You can experimentally determine that once it is dissolved, the water molecules present in the crystal lattice are gone into the solution.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Which type of bond is stronger ionic or covalent class 12 chemistry CBSE
What organs are located on the left side of your body class 11 biology CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
When people say No pun intended what does that mea class 8 english CBSE