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# 0.553 g of a boron - hydrogen compound created a pressure of 0.658 atm in a bulb of 407 mL at $100^{\large\circ}C$. Analysis showed it to $85.7\%$ boron . Calculate its molecular formula.

$(a)\;B_4H_6\qquad(b)\;B_3H_8\qquad(c)\;B_5H_7\qquad(d)\;B_5H_9$

For B-H compound
P = 0.658 atm
$V = \large\frac{407}{1000}$ litre
T = 373 K
w = 0.553 g
$PV = \large\frac{w}{m}RT$
$0.658\times\large\frac{407}{1000} = \large\frac{0.553}{m}\times0.0821\times373$
m = 63.23
100 g compound has 85.7 g of B
63.23 g compound has = $\large\frac{85.7\times63.23}{100}$ g of B
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = 54.19 \;g \;of \;B$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\large\frac{54.19}{10.8}$ g atom of B.
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = 5$ g of atom of B
Formula becomes $B_5H_x$
$5\times10.8 + x = 63.25$
$x = 9.25$
$x = 9 (an\; integer)$
Hence formula of the compound is $B_5H_9$