49th question?3g of a metal M combines with 2 g of oxygen to form metal oxide
Answers
Answer:
Explanation:equivalent means Normality×volume in liters=0.03=normality×0.5=0.03/0.5=0.06,
NormaAlity=molarity×nfactor,,,0.06=0.01 × n factor nfactor is 6
We assume that nfactor 6 it must be aluminium tri oxide,accto limiting reagent,the limiting reagent is metal 4moles ofmetal that react with 3moles oxygen to form 2moles metal tri oxide
Answer: Molar Mass is 120.
Explanation:
3 g of Metal combined with 2 g of Oxygen.
Number of Equivalent of Metal = Number of Equivalent of Oxygen
=> Weight of Metal/Equivalent Weight of Metal = Weight of Oxygen/Equivalent Weight of Oxygen
=> 3/x = 2/8
[Assuming Equivalent Weight of Metal to be x ;
Equivalent Weight of Oxygen= 8]
=> x = 12
Equivalent Weight of Metal Oxide = Equivalent Weight of Metal + Equivalent Weight of Oxygen
= 12 + 8 =20
Number of Equivalent of Metal Oxide = Weight of Metal Oxide/ Equivalent Weight of Metal Oxide
=> 0.03 = Weight of Metal Oxide/20
=> Weight of Metal Oxide = 0.6
Molarity = (Weight of Metal Oxide × 1000)/( Molecular weight of Metal Oxide × Volume of Solution)
=> 0.01 = (0.6 ×1000)/(Molecular weight of Metal Oxide×500)
=> Molecular weight of Metal Oxide= 120 (Ans.)