Question

A compound of phosphorus and oxygen contains 43.64 % by mass phosphorous. The empirical formula is (a) PO (6) PO₂ (c) P₂0 (d) PO (1) P₃0. Assuming ideal gas behavior, the density, in g/L of N,O(g) at 1.53 atm and 45.2 °C is (b) 1.76 g/L (c) 16.4 g/L (a) 18.2 g/L (e) none of the above. (d) 2.58 g/L When the measured quantity 7.43 g/cm is converted to kg/m the answer is (a) 7.43 x10 kg/m² (b) 7.43 x 10 kg/m (c) 7.43 kg/m? (d) 7.43 x 10 kg/m (e) none of the previous Hint: Use V/8 formula. The number of lone pairs about the central Xe in the Lewis dot structures of XOF. XeF, are respectively (a) 1,3 (b) 0,2 (c) 0,,3 (d) 1,2 (e) none of the previous

Answer 1

Answer 5):- P = 43.64% ; O = (100 - 43.64)% = 56.36%

Elemen Atomic 70 I Simplest Mass (g/mol) Atomic Ratio fatio I MAS P. 2 143.64 30.973 43.64 1.409 30.973 1.409 T.409 56.36 15.999 56.36 3.523 13.999 1.409 = 3.523 2.50

Empirical formula = P₂O₅

Option C) is correct one.

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_{​​​​​​}Answer 6):- According to Ideal gas equation,

PV = nRT

As, n = m/M

PV = (m/M) RT

PM = (m/V) RT

PM = dRT ..........(1) (d= mass/volume = m/V)

Where;

P = Pressure (in atm) ,

V = Volume ( in L),

R = gas constant = 0.0821 L-atm/K-mol

T = Temperature (in Kelvins)

n = number of moles

M = molar mass in g/mol

d = density (g/L)

Given:-

P = 1.53 atm ;

T = 45.2 ° C = 318.2 K ;

Molar mass (M) of N​​​​​​​₂O = 44.013 g/mol;

R = 0.0821 L-atm/K-mol;

d = ?

Using equation (1),

1.53 × 44.013 = d × 0.0821 × 318.2

On solving, we get,

d = 2.578 g/L

Hence density of N₂O = 2.58 g/L

Option (d) is correct one.

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Answer 7):- 1 kg = 1000

​​​​​1g = 1/1000 kg = 0.001 kg

1 m = 100 cm

1 cm = 1/100 m = 0.01 m

7.43 g/cm​​​​​​​² = 7.43 × (0.001kg) / (0.01m)² = 7.43 × 0.001 kg / 0.0001 m² = 74.3 kg/m² =

7.43 × 10¹ kg/m²

Option (d) is correct one.

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Answer 8):- XeOF4 = 1 lone pair

XeF2 = 3 lone pairs

Option (a)- 1,3 is correct answer.

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