JEE MAIN - Chemistry (2021 - 1st September Evening Shift - No. 19)
If 80 g of copper sulphate CuSO4 . 5H2O is dissolved in deionised water to make 5L of solution. The concentration of the copper sulphate solution is x $$\times$$ 10$$-$$3 mol L$$-$$1. The value of x is _____________.
[Atomic masses Cu : 63.54u, S : 32u, O : 16u, H : 1u]
[Atomic masses Cu : 63.54u, S : 32u, O : 16u, H : 1u]
Answer
64
Explanation
Given, mass of CuSO4 . 5H2O = 80 g
The concentration of copper sulphate solution is x $$\times$$ 10$$-$$3 mol/L.
Molarity = $${{Number\,of\,moles\,of\,solute} \over {Volume\,of\,solution(L)}}$$ ..... (i)
Molar mass of CuSO4 . 5H2O = 63.54 + 32 + 16 $$\times$$ 4
= 5 $$\times$$ 18 = 249.54 g/mol
Number of moles of solute = $${{Weight\,of\,solute} \over {Molecular\,mass\,of\,solute}}$$
= $${{80g} \over {249.54g/mol}}$$ = 0.32 mol
Volume of solution = 5 L
From Eq. (i),
Molarity = $${{0.3205} \over 5}$$ = 64.11 $$\times$$ 10$$-$$3 mol/L
$$\therefore$$ x = 64.11
or x $$\approx$$ 64
Hence, answer is 64.
The concentration of copper sulphate solution is x $$\times$$ 10$$-$$3 mol/L.
Molarity = $${{Number\,of\,moles\,of\,solute} \over {Volume\,of\,solution(L)}}$$ ..... (i)
Molar mass of CuSO4 . 5H2O = 63.54 + 32 + 16 $$\times$$ 4
= 5 $$\times$$ 18 = 249.54 g/mol
Number of moles of solute = $${{Weight\,of\,solute} \over {Molecular\,mass\,of\,solute}}$$
= $${{80g} \over {249.54g/mol}}$$ = 0.32 mol
Volume of solution = 5 L
From Eq. (i),
Molarity = $${{0.3205} \over 5}$$ = 64.11 $$\times$$ 10$$-$$3 mol/L
$$\therefore$$ x = 64.11
or x $$\approx$$ 64
Hence, answer is 64.
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