Question

A quantity of $$1\ g$$ of metallic carbonate $$XCO_3$$ is completely converted into a chloride. $$XCl_2$$ weighing $$1.11\ g$$. The atomic mass of the element '$$X$$' is ____________.

A
10
B
20
C
30
D
40

Solution

The correct option is D $$40$$$$XCO_3 \longrightarrow XCl_2$$$$1\ gm\quad \quad 1.11\ gm$$$$\frac { mole\ of\ { XCO }_{ 3 }\\ \downarrow \\ 1 }{ M+12+48 } \begin{matrix} \longrightarrow \\ = \end{matrix}\frac { mole\ of\ { XCl }_{ 2 }\\ \downarrow \\ 1.11 }{ M+71 }$$where, $$M=$$ molar mass of $$X$$.Now,$$M+71=(M+60)1.11$$$$M+71=1.11\ M+66.6$$$$4.4=0.11\ M$$$$M=40$$Option $$(D).$$Physics

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