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1151.

`4.2 g` of a metallic carbonate `MCO_(3)` was heated in a hard glass tube and `CO_(2)` evolved was found to have `1120 mL` of volume at `STP`. The `Ew` of the metal isA. 12B. 24C. 18D. 15

Answer» Correct Answer - A
`22400 mL = 1 "mol of" CO_(2) = 2 Eq "of" CO_(2)`
`11200 mL = 1//2 "mol of" CO_(2)` at `STP` would be is `Ew`.
`Ew` of metallic carbonate `= (4.2 xx 11200)/(1120) = 42 g`
`Ew` of metal `= Ew "of" MCO_(3) - Ew "of" CO_(3)^(2-)`
`= 42 - 30 = 12`
`[Ew "of" CO_(3)^(2-) = (60)/(2) = 30]`
Alternatively:
`22400 mL = 1 "mol of" CO_(2) = 2 Eq CO_(3)^(2-)`
`11200 mL implies 1 Eq "of" CO_(3)^(2-)`
`: 1120 mL implies (1)/(11200) xx 1120 = 0.1 Eq "of" CO_(3)^(2-)`
Eq of `MCO_(3) = Eq "of" CO_(3)^(2-)`
`("weight")/(Ew) = ("Weight")/(Ew) = 0.1 Eq`
`:. 0.1 = (4.2)/(Ew "of" MCO_(3))`
`:. 0.1 Eq "of" MCO_(3) = 42`
`:. Ew "of" M = Ew "of" MCO_(3) - Ew "of" CO_(3)^(2-)`
`= 42 - 30 = 12`
Discussion
1152.

A mixture of formic acid and oxalic acid is heated with conc. `H_(2)SO_(4)`. The gas producted is collected and treated with `KOH` solution, whereby the volume decrease by `1//6th`. The molar ratio of the two acids (formic acid/oxalic acid) isA. `4:1`B. `1:4`C. `2:1`D. `1:2`

Answer» Correct Answer - A
Let `x` mol of `HCOOH` and `y` mol of `COOH)_(2)`.
`underset(x)(HCOOH) overset(H_(2)SO_(4))rarr H_(2) + underset(x)(CO)`
`underset(x)((COOH)_(2)) overset(H_(2)SO_(4))rarr H_(2)O + underset(x)(CO) + underset(y)(CO_(2))`
Total moles of `(CO + CO_(2)) = x + 2y`
Total moles of `CO_(2) = y`
According to the question:
`(x + 2y) xx (1)/(6) = y`
`:. (x)/(y) = 4:1`
Alternatively:
Mole fraction of `CO_(2) = (y)/(x + 2y) = (1)/(6)`
`:. (x)/(y) = 4:1`
Discussion
1153.

Solution containing `23 g HCOOH` is/are :A. `46 g` of `70% ((w)/(v)) HCOOH (d_("solution") = 1.40 g//mL)`B. `50 g` of `10 M HCOOH (d_("solution") = 1 g//mL)`C. `50 g` of `25% ((w)/(w)) = HCOOH`D. `46 g` of `5" " M" "HCOOH (d_("solution") = 1 g//mL)`

Answer» Correct Answer - A::B
`(a)` In `100mL(140 g)` solution mass of solute
`= 70= (70)/(140)xx46=23 g`
`(b) 10M=("Mass of solute" //46)/((50)/(1xx1000))`
Mass of solute `=23 g`
`(c ) 100 g` solution contain `25 g` of solute mass of solute `=(25)/(100)xx50=12.5`
`(d) 5 M =("Mass of solute"//46)/(46//1000)`
Mass of solute `=10.58 g`
Discussion