Home
Class 12
CHEMISTRY
A solution containing 28 g of phosphorus...

A solution containing 28 g of phosphorus in 315 g `CS_(2)(b.p. 46.3^(@)C`) boils at `47.98^(@)C` . If `K_(b)` for `CS_(2)` is `2.34` K kg `mol^(-1)` . The formula of phosphorus is (at , mass of P = 31).

A

`P_(6)`

B

`P_(4)`

C

`P_(3)`

D

`P_(2)`

Text Solution

Verified by Experts

The correct Answer is:
B
Promotional Banner

Topper's Solved these Questions

  • LIQUID SOLUTIONS

    GRB PUBLICATION|Exercise 24|8 Videos
  • LIQUID SOLUTIONS

    GRB PUBLICATION|Exercise 26|7 Videos
  • LIQUID SOLUTIONS

    GRB PUBLICATION|Exercise 8|12 Videos
  • IONIC EQUILIBRIUM

    GRB PUBLICATION|Exercise All Questions|526 Videos
  • METALLUGY

    GRB PUBLICATION|Exercise Subjective type|1 Videos

Similar Questions

Explore conceptually related problems

A solution containing 28 g of phosphorus in 315 g CS_(2)(b.p.46.3^(@)C boils at 47.98^(@)C . If k_(b) for CS_(2) is 2.34 K kg mol^(-1) . The formula of phosphorus is (at .massof P= 31 ).

A solution containing 3.3g of a substance in 125g of benzne (b.pt = 80^(@)C ) boils at 80.66^(@)C . If K_(b) for benzene is 3.28 "K kg mol"^(-1) the molecular mass of the substance will be :

An aqueous solution of glucose containing 12 g in 100 g of water was found to boil at 100.34^(@)C . Calculate of K_(b) for water in K mol^(-1)Kg .

The density of phosphorus vapour at 310^(@)C and 775 torr is 2.64 g dm^(-3) . What is the molecular formula of phosphorus ?

3.795 g of sulphur is dissolved in 100g of CS_(2) . This solution boils at 319.81 K. What is the molecular formula of sulphur in solution? The boiling point of CS_(2) is 319.45 kJ (Given that K_(b) for CS_(2) = 2.42 K kg mol^(-1) and atomic mass of S = 32)