1./ An element X has a tribromide with the empirical formula XBr3 and a trichlor

Question

1./ An element X has a tribromide with the empirical formula XBr3 and a trichloride with the empirical formula XCI3. The tribromide is converted to the trichloride according to the equation XBr3 + cizxcI3 + Br2 If the complete conversion of 1.579 g of XBr3 results in the formation of 0.855 g of XC13, what is the atomic mass of the element X? Atomic mass X g mol-1 2. A sample of a substance with the empirical formula XBr2 weighs 0.4071 g. When it is dissolved in water and all its bromine is converted to insoluble AgBr by addition of an excess of silver nitrate, the mass of the resulting AgBr is found to be 0.8305 g. The chemical reaction is XBr2+ 2 AgNO32AgBr + X(NO3)2 (a) Calculate the formula mass of XBr2. Formula mass XBr2 g mol-1 (b) Calculate the atomic mass of X. Atomic mass x - g mol-1

Explanation / Answer

molar mass of Cl = 35.5 g/mol
molar mass of Br = 79.9 g/mol

let m1 and m2 be molar mass of XBr3 and XCl3 respectively

number of mol = (given mass)/(molar mass)
number of mol of XBr3 = 1.579/m1
number of mol of XCl3 = 0.855/m2

reaction taking place is
2XBr3 + 3Cl2 --> 2XCl3 + 3Br2

according to reaction
number of mol of XBr3 reacted = number of mol of XCl3 formed
so,
1.579/m2 = 0.855/m1
m1/m2 = 0.542

let x be the atomic mass of X
also,
molar mass of XCl3 = 3*(atomic mass of Cl) + (atomic mass of X)
= (3*35.5) + x

= 106.5 + x
molar mass of XBr3 = 3(atomic mass of Br) + (atomic mass of X)
= (3*79.9) + x

= 239.7 + x

so,
m1/m2 = (35.5+x)/(79.9+x)
0.542 = (106.5+x)/(239.7+x)
0.542*(239.7 + x) = 106.5 + x
129.9 + 0.542*x = 106.5 + x
0.458*x = 23.4
x = 51.1 g/mol

Answer : 51.1 g/mol

i am allowed to do only one question at a time

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