SATHEE: Some Basic Concepts Of Chemistry Question 85

Some Basic Concepts Of Chemistry Question 85

Question: 1 g mixture of equal number of mole of $L i_{2} C O_{3}$ and other metal carbonate $(M_{2} C O_{3})$ required 21.6 mL of $0.5 N H C l$ for complete neutralization reaction. What is the approximate atomic mass of the other metal?

Options:

A) 25

B) 23

C) 51

D) 118

Show Answer

Answer:

Correct Answer: D

Solution:

[d] Let x g of $L i_{2} C O_{3}$ and $(1 - x) g$ of $M_{2} C O_{3}$ present in given mixture $2 \times$ Total moles of carbonates = moles of HCl and $2 \times (\frac{x}{74} + \frac{1 - x}{2 M + 60}) = 21.6 \times 0.5 \times 10^{- 3}$ $x = 0.20$
$∴ \frac{x}{74} = \frac{1 - x}{2 M + 60}$ $M = 118$

पिछला

अगला

गोपनीयता नीति

द्वारा संचालित Prutor@IITK

sathee@iitk.ac.in

SATHEE का मीडिया कवरेज

खेल स्टोर

ऐप स्टोर

Ask SATHEE

Welcome to SATHEE !
Select from 'Menu' to explore our services, or ask SATHEE to get started. Let's embark on this journey of growth together! 🌐📚🚀🎓

I'm relatively new and can sometimes make mistakes.
If you notice any error, such as an incorrect solution, please use the thumbs down icon to aid my learning.
To begin your journey now, click on "I understand".

Some Basic Concepts Of Chemistry Question 85

Question: 1 g mixture of equal number of mole of Li2CO3 and other metal carbonate (M2CO3) required 21.6 mL of 0.5NHCl for complete neutralization reaction. What is the approximate atomic mass of the other metal?

Options:

Answer:

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Menu

Question: 1 g mixture of equal number of mole of $L i_{2} C O_{3}$ and other metal carbonate $(M_{2} C O_{3})$ required 21.6 mL of $0.5 N H C l$ for complete neutralization reaction. What is the approximate atomic mass of the other metal?