1.

The percentage of an element M is 53 in its oxide of molecular formula M_(2)O_(3).Its atomic mass is about

Answer»

45
9
18
27

Solution :If m is the atomic mass of the element M, then
`2m+48G" of "M_(2)O_(3)" contain O"=48g.`
`therefore %" of O in "M_(2)O_(3)=(48)/(2m+48)xx100=53"(Given)"`
`therefore""2m+48=(4800)/(53)=90.56`
`"or"2m=42.56"or"m=26.28~~27`


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