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)

${\mathrm{MBr}}_{\mathrm{x}}+\mathrm{XHCl}\to {\mathrm{MCl}}_{\mathrm{x}}+\mathrm{XHBr}$

For the given reaction:
Equivalent of MBr_x​ =  Equivalent of MCl_x​

$\frac{1.878 }{\mathrm{E}+80}=\frac{1.0}{ \mathrm{E}+35.5}$

Equivalent mass of the metal (E)=15.18

Now, atomicmass×specificheat≃6.4

Atomic mass of metal = $\frac{ 6.4}{0.14}=45.71$

Valency of metal = $\frac{\mathrm{Atomic} \mathrm{mass}}{\mathrm{Eq}.\mathrm{mass}}=\frac{45.71}{15.18}=3.01=3(\because \mathrm{integer})$

Exact atomic mass of metal=15.18×3=45.54

Molar mass of  MBr_x​ or MBr_3​(x=3)=45.54+80×3=285.54gmol⁻¹

So, the nearest integer is 286.