Question

a boron hydrogen compound weighing 0.0553 g created a pressure of 0.658 atm in a bulb of 40.7 ml volume at 100 degree celcius. Analysis showed it to be 85.7% boron. What is its molecular formula?

Answer 1

For B-H compound

P = 0.658 atm

V=4071000V=4071000 litre

T = 373 K

w = 0.553 g

PV=wmRTPV=wmRT

0.658×4071000=0.553m×0.0821×3730.658×4071000=0.553m×0.0821×373

m = 63.23

100 g compound has 85.7 g of B

63.23 g compound has = 85.7×63.2310085.7×63.23100 g of B

=54.19gofB=54.19gofB

=54.1910.8=54.1910.8 g atom of B.

=5=5 g of atom of B

Formula becomes B5HxB5Hx

5×10.8+x=63.255×10.8+x=63.25

x=9.25x=9.25

x=9(aninteger)x=9(aninteger)

Hence formula of the compound is B5H9

Answer 2

Answer:

Explanation:

63.23 g compound has = 85.7×63.2310085.7×63.23100 g of B

=54.19gofB

=54.1910.8

5 g of atom of B

Formula becomes B5Hx

5×10.8+x=63.25

x=9.25

x=9

Hope it will help you