10 g of `MgCO_(3)` decomposes on heating to 0.1 mole `CO_(2)` and 4 g MgO. The percent purity of `MgCO_(3)` is

Options

A. 0.24
B. 0.44
C. 0.54
D. 0.84

Answer

C

Explanation

Since `MgCO_(3) overset(triangle)rarr MgO+CO_(2)(g)` Mass ratio (ing ) 84:40:44
Given values 10 g 4g 1 mol
`(=44g){because n=(w)/(m)}`
(i) `because 84 g of MgCO_(3)` give MgO=40g
Thus Mgo obtained is less by
4.76 -4.0 =0.76 g
(ii) `because 84 g of MgCO_(3)` given `CO_(2)=44g`
`therefore` 10 g of `MgCO_(3)` give `CO_(2)=(44xx10)/(84)=5.23 g`
Thus `CO_(2)` obained is less by 5.23 -4.4 =0.84 g
`therefore` 100 g of `MgCO_(3)` give less products by =1.600 g
Hence `MgCO_(3)=10-16=84%`