If 0.44 g of a colourless oxide of nitrogen occupies 224 ml at 1520 mm

1.	If 0.44 g of a colourless oxide of nitrogen occupies 224 ml at 1520 mm Hg and 273 °C,then the compound is :(a) N₂O (6) NO₂(c) NO4 (d) N2O2 use the formula pv= nrt and solve
Answer» Answer: (a) Explanation: According to the ideal gas EQUATION:' PV=nRTPV=nRT P= Pressure of the gas = 1520 mmHg = 2 atm V= Volume of the gas = 224 mL = 0.224 L T= Temperature of the gas = 273°C=(273+273)K=546 K (0°C = 273 K) R= Value of gas constant = 0.0821 Latm/K mol n=\frac{PV}{RT}=\frac{2\times 0.225L}{0.0821 \times 546}=0.010molesn= RT PV = 0.0821×546 2×0.225L =0.010moles To calculate the MOLES, we use the equation: \text{Number of moles}=\frac{\text{Given MASS}}{\text {Molar mass}}Number of moles= Molar mass Given mass 0.010={0.44}{Molar mass}}0.010= Molar mass 0.44 {\text {Molar mass}}=44gMolar mass=44g

If 0.44 g of a colourless oxide of nitrogen occupies 224 ml at 1520 mm Hg and 273 °C,then the compound is :(a) N₂O (6) NO₂(c) NO4 (d) N2O2 use the formula pv= nrt and solve

Answer»

Answer:

(a)

Explanation:

According to the ideal gas EQUATION:'

PV=nRTPV=nRT

P= Pressure of the gas = 1520 mmHg = 2 atm

V= Volume of the gas = 224 mL = 0.224 L

T= Temperature of the gas = 273°C=(273+273)K=546 K (0°C = 273 K)

R= Value of gas constant = 0.0821 Latm/K mol

n=\frac{PV}{RT}=\frac{2\times 0.225L}{0.0821 \times 546}=0.010molesn=

0.0821×546

2×0.225L

=0.010moles

To calculate the MOLES, we use the equation:

\text{Number of moles}=\frac{\text{Given MASS}}{\text {Molar mass}}Number of moles=

Molar mass

Given mass

0.010={0.44}{Molar mass}}0.010=

Molar mass

0.44

{\text {Molar mass}}=44gMolar mass=44g