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

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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 $B_5H_9$

Hence answer is (D)

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