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Standard XII

Chemistry

Percentage Composition

A Zn rod weig...

Question

A Zn rod weighing 25 g was kept in 100mL of 1 M $C u S O 4$ solution. After a certain time, the molarity of $C l^{2 +}$ in solution was 0.8. What was the molarity of ${S O}_{4}^{2 -}$ ? What was the mass of the $Z n$ rod after cleaning?
(The atomic mass of $Z n = 65.4$ )

23.692 g

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33.692 g

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43.692 g

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13.692 g

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Solution

The correct option is C $23.692 g$
Eq. of $C u^{2 +}$ before reaction $= 100 \times \frac{1}{1000} \times 2 = 0.2$

Eq. of $C u^{2 +}$ after reaction $= 100 \times \frac{0.8}{1000} \times 2 = 0.16$

$∴$ Eq. of $C u^{2 +}$ lost $= 0.2 - 0.16 = 0.04$

$∴$ Eq. of $Z n$ lost $= 0.04$

$∴$ $w_{Z n} / E_{Z n} = 0.04$ or $w_{Z n} = 0.04 \times \frac{65.4}{2} = 1.308 g$

Thus, net mass of $Z n$ rod $= 25 - 1.308 = 23.692 g$

Since, reactions are $Z n \to Z n^{2 +} + 2 e$

$∴$ No, change in molarity of $S O_{4}^{2 -}$ ,i.e., $1 M$

Hence, the correct option is $A$

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A Zn rod weighing 25 g was kept in 100mL of 1 M CuSO4 solution. After a certain time, the molarity of Cl2+ in solution was 0.8. What was the molarity of SO2−4? What was the mass of the Zn rod after cleaning? (The atomic mass of Zn=65.4)

A Zn rod weighing 25 g was kept in 100mL of 1 M $C u S O 4$ solution. After a certain time, the molarity of $C l^{2 +}$ in solution was 0.8. What was the molarity of ${S O}_{4}^{2 -}$ ? What was the mass of the $Z n$ rod after cleaning?
(The atomic mass of $Z n = 65.4$ )