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Standard IX

Physics

Density

100 mL of 0...

Question

$100 m L$ of $0.1 n$ hypo decolourised iodine by the addition of $x g$ of crystalline copper sulphate to excess of $K l$ .The value of $x$ is:

5.0 g

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1.25 g

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2.5 g

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4 g

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Solution

The correct option is C $2.5 g$
100 ml of 0.1 hypo $(N a_{2} S_{2} O_{3})$ = 100 ml of 0.1 N (Normality) $C u S O_{4} .5 H_{2} O$ = $249.5 \times 0.1) / (1000)$ $\times 100 = 2.5 g$

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Similar questions

1.2475

g of crystalline copper sulphate was dissolved in water and excess of KI was added. The liberated iodine consumed

50

\frac{N}{10} N a_{2} S_{2} O_{3}

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Iodine liberated was titrated against

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10

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C u S O_{4} + 2 K I ⟶ K_{2} S O_{4} + C u I_{2}

2 C u I_{2} ⟶ C u_{2} I_{2} + I_{2}

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100 mL of 0.1 nhypo decolourised iodine by the addition of x g of crystalline copper sulphate to excess of Kl.The value of x is:

The correct option is C 2.5 g100 ml of 0.1 hypo (Na2S2O3) = 100 ml of 0.1 N (Normality) CuSO4.5H2O = 249.5×0.1)/(1000) ×100=2.5g

$100 m L$ of $0.1 n$ hypo decolourised iodine by the addition of $x g$ of crystalline copper sulphate to excess of $K l$ .The value of $x$ is:

The correct option is C $2.5 g$
100 ml of 0.1 hypo $(N a_{2} S_{2} O_{3})$ = 100 ml of 0.1 N (Normality) $C u S O_{4} .5 H_{2} O$ = $249.5 \times 0.1) / (1000)$ $\times 100 = 2.5 g$