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Question: The cupric salt \((C{{u}^{+2}})\) of a monobasic acid contains 3 molecules of water of hydration per...

The cupric salt (Cu+2)(C{{u}^{+2}}) of a monobasic acid contains 3 molecules of water of hydration per atom of Cu. 1g of hydrated salt on strong heating yields 0.3306g of CuO. The equivalent mass of acid is (as the nearest integer):

Explanation

Solution

The equivalent weight of a substance is defined as the number of parts weight of substance which will combine with or displace directly or indirectly 1.008 parts by weight to hydrogen, 8 parts of oxygen, and 35.5 parts of chlorine by weight are the equivalent parts by weight of another element.

Complete step by step answer:
In the reaction, the equivalent weight depends on the type of reaction involved in it. Due to the equivalent weight is a relative quantity, so it is unitless.
If the equivalent weight expressed in grams is known as Gram Equivalent Weight (GEW).
The ratio of atomic weight and n-factor is known as equivalent weight.
Equivalent weight (E) = atomic weightn−factor\dfrac{atomic\text{ }weight}{n-factor}
For acid or base, the n-factor is acidity/basicity, which is the number of H+{{H}^{+}} ion or number of dissociable OH−O{{H}^{-}} ions.
The equivalent weight of acid = molecular weight of an acidbasicity of an acid\dfrac{molecular\text{ }weight\text{ }of\text{ }an\text{ }acid}{basicity\text{ }of\text{ }an\text{ }acid}

Let the equivalent weight of acid is ‘x’, CuA2.3H2OCu{{A}_{2}}.3{{H}_{2}}O which represents the cupric salt of a monobasic acid contains 3 molecules of water of hydration per atom of Cu.
Equivalent weight of acid = 63.55+2(x−1)+3X18=115.25+2x63.55+2(x-1)+3X18=115.25+2x-- (1)
CuA2.3H2O→CuOCu{{A}_{2}}.3{{H}_{2}}O\to CuO
Molecular mass of CuO =79.5g
Moles of CuO = 0.330679.5=4.156X10−3mol\dfrac{0.3306}{79.5}=4.156X{{10}^{-3}}mol
Molar mass of CuO =wt4.156X10−3mol=240.6g/mol\dfrac{wt}{4.156X{{10}^{-3}}mol}=240.6g/mol -- (2)
Since, (1) = (2)
115.5+2x =240.6g/mol
X =62.58
Hence the equivalent weight of anhydrous acid = 62.58g

Note: There are many methods for determination of equivalent weight by hydrogen replacement, by oxide formation, by metal chloride formation, by metal displacement, and by electrolysis. For oxidizing or reducing agents, the n-factor is the number of moles of electrons gained or lost per mole of oxidizing or reducing agents respectively.