Notifications
Clear all
A transition metal M forms a volatile chloride which has a vapour density of 94.8. If it contains 74.75% of chlorine the formula of the metal chloride will be
Some Basic Concepts of Chemistry
1
Posts
2
Users
0
Likes
469
Views
0
12/08/2021 11:52 am
Topic starter
A transition metal M forms a volatile chloride which has a vapour density of 94.8. If it contains 74.75% of chlorine the formula of the metal chloride will be
(a) MCl3
(b) MCl2
(c) MCl4
(d) MCl5
Answer
Add a comment
Add a comment
Topic Tags
1 Answer
0
12/08/2021 12:06 pm
Correct answer: (c) MCl4
Explanation:
74.75% of chlorine means 74.75g chlorine is present in 100g of metal chloride.
Weight of metal = 100g – 74.75g
= 25.25g
Equivalent weight
= \(\frac{weight\;of\;metal}{weight\;of\;chlorine}\) x 35.5
= \(\frac{25.25}{74.75}\) x 35.5 = 12
Valency of metal
= \(\frac{2 \times V.D.}{Equivalent\;wt. of\;metal + 35.5}\)
= \(\frac{2 \times 94.8}{12 + 35.5}\) = 4
∴ Formula of compound = MCl4
Add a comment
Add a comment
Forum Jump:
Related Topics
-
2.76 g of silver carbonate (at mass of Ag 108) on being heated strongly yield a residue weighing
3 years ago
-
12 g of Mg (atomic mass 24) will react completely with hydrochloric acid to give
3 years ago
-
If potassium chlorate is 80% pure, then 48 g of oxygen would be produced from (atomic mass of K = 39)
3 years ago
-
When 2.5 g of a sample of Mohr's salt reacts completely with 50 mL of N/10 KMnO4 solution. The % purity of the sample of Mohr's salt is:
3 years ago
-
1 g mixture of equal number of mole of Li2CO3 and other metal carbonates (M2CO3) required 21.6 mL of 0.5N HCl for complete neutralisation reaction. What is the approximate atomic mass of the other metal?
3 years ago
Forum Information
- 321 Forums
- 27.3 K Topics
- 53.8 K Posts
- 1 Online
- 12.4 K Members
Our newest member: Stripchat
Forum Icons:
Forum contains no unread posts
Forum contains unread posts
Topic Icons:
Not Replied
Replied
Active
Hot
Sticky
Unapproved
Solved
Private
Closed