Calculate the molecular mass of potassium permanganate, ${KMnO}_{4}$ .
[Atomic masses of K=39,Mn=55,O=16]

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Solution

Molecular mass of ${KMnO}_{4}$
$= 1 \times Atomic mass of K + 1 \times Atomic mass of Mn + 4 \times Atomic mass of O = 1 \times 39 + 1 \times 55 + 4 \times 16 = 39 + 55 + 64 = 94 + 64 = 158 u$
Therefore,the molecular mass of ${KMnO}_{4}$ is 158u.

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Similar questions

When heated, potassium permanganate decomposes according to the following equation:
2KMnO₄ $\overset{}{\to}$ K₂MnO₄ + MnO₂ + O₂
(a) Some potassium permanganate was heated in a test tube. After collecting one litre of oxygen at room temperature, it was found that the test tube had undergone a loss in mass of 1.32 g. If one litre of hydrogen under the same conditions of temperature and pressure has a mass of 0.0825 g, calculate the relative molecular mass of oxygen.
(b) Given that the molecular mass of potassium permanganate is 158 g, what volume of oxygen (measured at room temperature) would be obtained by the complete decomposition of 15.8 g of potassium permanganate? (Molar volume at room temperature is 24 litres). (K = 39, Mn = 55, O = 16)

(a) A flask contains 3.2 g of sulphur dioxide. Calculate the following:
(i) The moles of sulphur dioxide present in the flask.
(ii) The number of molecules of sulphur dioxide present in the flask.
(iii) The volume occupied by 3.2 g of sulphur dioxide at STP. (S = 32, O = 16)

(b) The reaction of potassium permanganate (VII) with acidified iron (II) sulphate is given below:
2KMnO₄ + 10FeSO₄ + 8H₂SO₄ $\overset{}{\to}$ K₂SO₄ + 2MnSO₄ + 5Fe₂(SO₄)₃ + 8H₂O
If 15.8 g of potassium permanganate (VII) was used in the reaction, calculate the mass of iron (II) sulphate used in the above reaction. (K = 39, Mn = 55, Fe = 56, S = 32, O = 16)

Calculate the relative molecular masses of (Use K = 39, Cl = 35.5, O = 16, C = 12, H = 1, Na = 23, N = 14, S = 32)
(a) potassium chlorate
(b) sodium acetate
(c) chloroform
(d) ammonium sulphate

Q. Approximately how many moles of potassium permanganate,

K M n O_{4}

, are present in

40 g

of this compound? (Atomic weights:

K = 39; M n = 55; O = 16

)

Q. The reaction of potassium permanganate with acidified iron [II] sulphate is given below :-

2 K M n O_{4} + 10 F e S O_{4} + 8 H_{2} S O_{4} \to K_{2} S O_{4} + 2 M n S O_{4} + 5 F e_{2} (S O_{4})_{3} + 8 H_{2} O

15.8

g of potassium permanganate was used in the reaction, calculate the mass of iron [II] sulphate used in the above reaction.
[Molecular Weight of

K = 39, M n = 55, F e = 56, S = 32, O = 16

g/mole]

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Calculate the molecular mass of potassium permanganate,KMnO4.[Atomic masses of K=39,Mn=55,O=16]

Molecular mass of KMnO4=1×AtomicmassofK+1×AtomicmassofMn+4×AtomicmassofO=1×39+1×55+4×16=39+55+64=94+64=158uTherefore,the molecular mass of KMnO4is 158u.

Calculate the molecular mass of potassium permanganate, ${KMnO}_{4}$ .
[Atomic masses of K=39,Mn=55,O=16]

Molecular mass of ${KMnO}_{4}$
$= 1 \times Atomic mass of K + 1 \times Atomic mass of Mn + 4 \times Atomic mass of O = 1 \times 39 + 1 \times 55 + 4 \times 16 = 39 + 55 + 64 = 94 + 64 = 158 u$
Therefore,the molecular mass of ${KMnO}_{4}$ is 158u.