Question

0.84 gram of metal carbonate react with 40 ml of N/2 h2so4. Equivalent weight of the metal carbonate is​

Answer 1

Answer:

the equivalent weight of metal carbonate is 1.24 grams

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Answer 2

Answer:

Weight of the metal carbonate is 42 grams

Explanation:

We are given

Normality of $H_2So_4$ = 1/2M =  0.5M.

Now we'll find morality

$\text Normality=\text Molarity\times \text{Valence factor}$

(As we know that Valence factor of $H_2So_4$ = 2 )

$\text{Molarity of}H_2SO_4=\dfrac{0.5}{2}=0.25$

$Molarity=\dfrac{Moles}{\text{Volume (in mL)}}\times 1000\\\\ \text{Moles of }H_2SO_4=\dfrac{0.25\times 4}{1000}=0.01moles$

We can write moles as:

$\text Moles=\dfrac{\text{Given mass}}{\text{Molar mass}}\\\\\text{Molar mass of }H_2SO_4=98g/mol\\\\\text{Mass of }H_2SO_4=0.01mol\times 98g/mol=0.98g$

Equivalence of metal carbonate = Equivalence of sulfuric acid

$\text Equivalence=\dfrac{\text{Mass of the substance}}{\text{Equivalent weight of the substance}}$

$\text {Equivalent weight of } H_2SO_4=\dfrac{98}{2}=49$

Mass of sulfuric acid = 0.98 g

Let equivalent weight of metal carbonate be x grams

Mass of metal carbonate = 0.84 g

Now put the value in equation,

$\dfrac{0.98}{49}=\dfrac{0.84}{x}$

x = 42 grams

Weight of the metal carbonate is 42 grams