Question

7. Two oxides of a metal contains 50% and 40% metal (M), respectively. If the formula of first oxide is $\orange{ \sf \: MO_2}$, the formula of second oxide will be: $\sf \: (a) \: MO_2 \\ \sf (b) MO \\ \sf\ \textless \ br /\ \textgreater \ (c) MO_3 \\ \sf (d) M_2O_5$
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Answer 1

Answer :-

Formula of the 2nd Oxide is MO₃ [Option.c]

Explanation :-

It is given that the 1st oxide has 50% of metal "M" and formula is MO₂ .

Let the molar mass of metal M be M g/mol . Also, mass of Oxygen in 1st oxide is (16 × 2) = 32g.

⇒ M/(M + 32) × 100 = 50

⇒ 100 M = 50(M + 32)

⇒ 100 M = 50 M + 1600

⇒ 50 M = 1600

⇒ M = 1600/50

⇒ M = 32 g/mol

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For the 2nd Oxide, the percentage of metal "M" is 40 % .

Let the mass of Oxygen present be x grams. Again, putting values in percentage composition :-

⇒ 32/(32 + x) × 100 = 40

⇒ 3200 = 40(32 + x)

⇒ 3200 = 1280 + 40x

⇒ 40x = 1920

⇒ x = 1920/40

⇒ x = 48 g

Moles of Oxygen (in 2nd Oxide) :-

= Given Mass/Molar Mass

= 48/16

= 3

∵ There are 3 moles of Oxygen.

∴ Required formula is MO₃ .

Answer 2

Answer:

QUESTION :

• Two oxides of a metal contains 50% and 40% metal (M), respectively. If the formula of first oxide is $\orange{ \sf \: MO_2}$, the formula of second oxide will be: $\sf \: (a) \: MO_2 \\ \sf (b) MO \\ \sf\ \textless \ br /\ \textgreater \ (c) MO_3 \\ \sf (d) M_2O_5$

given :

• first oxide metal contain = 50 %

• second oxide metal contain = 40 %

to find :

• formula of second oxide will be = ?

• formula of second oxide will be = ?

solution :

first oxide, M = 50% , 0 = 50 %

formula for first oxide is = MO

molecular and mass is 16 + x where x is the mass of M

• second oxide = M = 40

• 0% = 60

• oxygen = 32/32+x = 0.5

• x= 32

• 16n /32+ 16 = 0.6

• n=3

• m ÷ 0

• 40 ÷ 60

• 4 ÷ 6

• 2÷ 3

• MO3

thus , the answer is MO3