Question

if 2.4 g of metal displace 1.12 litre hydrogen at normal temperature and pressure equivalent weight of metal would be ?​

Answer 1

Answer:  Equivalent mass of metal is 24 g.

Explanation:

Equivalent mass of a metal is defined as the mass which combines with 8 gram of oxygen or 35.5 grams of chlorine or 1 gram of hydrogen.

According to the ideal gas equation:'

$PV=nRT$

Pressure of the gas = 1 atm

Volume of the gas = 1.12 L

Temperature of the gas = 20°C = 293 K   (0°C = 273 K)

Value of gas constant in in kilopascals = 0.0821 Latm/K mol

$n=\frac{PV}{RT}=\frac{1atm\times 1.12L}{0.0821\times 293}=0.05mol$

Thus mass of hydrogen =$moles\times {\text {molar mass}}=0.05\times 2=0.1g$

Given:

0.1 gram of hydrogen combine with = 2.4 g of metal

So 1 gram of hydrogen combines with =$\frac{2.4}{0.1}\times 1= 24g$ of metal

Thus equivalent mass of metal is 24 g.

Answer 2

Explanation:

Equivalent mass of a metal is defined as the mass which combines with 8 gram of oxygen or 35.5 grams of chlorine or 1 gram of hydrogen.

According to the ideal gas equation:'

PV=nRTPV=nRT

Pressure of the gas = 1 atm

Volume of the gas = 1.12 L

Temperature of the gas = 20°C = 293 K (0°C = 273 K)

Value of gas constant in in kilopascals = 0.0821 Latm/K mol

n=\frac{PV}{RT}=\frac{1atm\times 1.12L}{0.0821\times 293}=0.05moln=

RT

PV

=

0.0821×293

1atm×1.12L

=0.05mol

Thus mass of hydrogen =moles\times {\text {molar mass}}=0.05\times 2=0.1gmoles×molar mass=0.05×2=0.1g

Given:

0.1 gram of hydrogen combine with = 2.4 g of metal

So 1 gram of hydrogen combines with =\frac{2.4}{0.1}\times 1= 24g

0.1

2.4

×1=24g of metal

Thus equivalent mass of metal is 24 g.

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