Question

0.4 g

of a metal was dissolved in

500 ml

of

(\frac{N}{10}) H_{2} {SO}_{4}

. When the reaction was over, the excess required was

150 ml (\frac{N}{15})

alkali for complete neutralization. Calculate the equivalent mass of metal.

Accepted Answer

Given that $0.4 g$ of a metal was dissolved in $500 ml$ of $(\frac{N}{10}) H_{2} {SO}_{4}$ . When the reaction was over, the excess required was $150 ml (\frac{N}{15})$ alkali for complete neutralization.

First, we should find out how much volume of $H_{2} {SO}_{4}$ has been reacted. We know that,

$N_{1} \times V_{1} = N_{2} \times V_{2} 0.1 \times V_{1} = \frac{1}{15} \times 0.15 V_{1} = 0.1 l = 100 ml remained$

The volume of acid reacted is $500 - 400 = 100 ml$

Number of moles of $H_{2} {SO}_{4}$ reacted is given by the formula,

$moles of acid = molarity \times volume of acid reacted moles of acid = (\frac{N}{eq . number}) \times 400 ml moles of acid = \frac{0.1}{n_{1}} \times 0.4 l = \frac{0.04}{n_{1}}$

Now, moles of metal participated is given by the formula,

$moles of metal = \frac{given mass}{molar mass} moles of metal = \frac{0.4}{molar mass} moles of metal = \frac{0.4}{equivalent mass \times n_{2}}$

But in a reaction, the product of a number of moles and the number of equivalents are the same. Hence,

$moles of acid \times n_{1} = moles of the metal \times n_{2} \frac{0.04}{n_{1}} \times n_{1} = \frac{0.4}{equivalent mass \times n_{2}} \times n_{2} Equivalent mass of metal = \frac{0.4}{0.04} = 10 g$

Odisha Board Solutions for Chapter: Redox Reaction, Exercise 3: QUESTIONS

Odisha Board Chemistry Solutions for Exercise - Odisha Board Solutions for Chapter: Redox Reaction, Exercise 3: QUESTIONS

Questions from Odisha Board Solutions for Chapter: Redox Reaction, Exercise 3: QUESTIONS with Hints & Solutions

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