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ChemistryMole ConceptExpetimental Determination of Atomic Mass/Molecuair massMedium2 minQB

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0.2 gm of metal oxide (MO) is treated with 50 ml of $\frac{N}{2} HCl$ solution. The excess of acid is neutralised by 30 ml . of $\frac{N}{2} NaOH$ solution. The atomic wt. of metal is

Options:

Answer:

Solution:

$\frac{0.2}{E} \times 1000 = (50 - 30) \times \frac{1}{2} = 10$

E = 20

Molecular weight = $20 \times 2 = 40$ .

Atomic weight of the metal = $40 - 16 = 24$ .

Stream:JEESubject:ChemistryTopic:Mole ConceptSubtopic:Expetimental Determination of Atomic Mass/Molecuair mass

⏱ 2mℹ️ Source: QB

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