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1.878g of MBr_x, when heated in a stream of HCl gas, was completely converted to chloride MCl_x, which weighed 1.0g. The specific heat of metal is 0.14cal/g. Then the molecular weight of metal bromide is (atomic weight of Br = 80)

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285.5

185.7

216

85.7

answer is A.

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Detailed Solution

Weight of MBr_X = 1.878g
Weight of MCl_x = 1g
EW of Cl^- = 35.5
Ew of Br^- = 80

\large \frac{{{W_{MB{r_x}}}}}{{{W_{MC{l_x}}}}}\; = \;\frac{{{E_{M\;}} + \;80}}{{{E_{M\;}} + \;35.5}}

\frac{{1.878}}{1}\; = \;\frac{{{E_{M\;}} + \;80}}{{{E_M}\; + \;35.5}}

1.878E_M + 66.7 = E_M + 80
0.878E_M = 13.3
E_M = 15.15
According to Dulong - Petit law
Specific heat x approximate Aw = 6.4

\large Approximate\;Aw\;=\;\frac{{6.4}}{0.14}\; = \;45.7

\large valency \;of\; the\; metal=\frac{45.7}{15.15}\simeq 3

Exact Aw of metal = 15.15 x 3 = 45.45

Formula of metal bromide is M⁺³ Br^- = MBr₃
MW of MBr₃ = 45.5+3(80)
= 285.5

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