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# 34.05 Ml of Phosphorus Vapour Weighs 0.0625 G at 546 °C and 0.1 Bar Pressure. What is the Molar Mass of Phosphorus? - CBSE (Science) Class 11 - Chemistry

ConceptIdeal Gas Equation Derivation of Ideal Gas Equation

#### Question

34.05 mL of phosphorus vapour weighs 0.0625 g at 546 °C and 0.1 bar pressure. What is the molar mass of phosphorus?

#### Solution 1

Given,

p = 0.1 bar

V = 34.05 mL = 34.05 × 10–3 L = 34.05 × 10–3 dm3

R = 0.083 bar dm3 K–1 mol–1

T = 546°C = (546 + 273) K = 819 K

The number of moles (n) can be calculated using the ideal gas equation as:

pV = nRT

=> n = (pV)/(RT)

= (0.1 xx 34.05 xx 10^(-3))/(0.083 xx 819)

= 5.01 xx 10^(-5)  mol

Therefore, molar mass of phosphorus = (0.0625)/(5.01 xx 10^(-5)) = 1247.5 g mol^(-1)

Hence, the molar mass of phosphorus is 1247.5 g mol–1.

#### Solution 2

Step 1 Calculation of volume at 0^@C and 1 bar pressure

(P_1V_1)/T_1 = (P_2V_2)/T_2   i.e (1xx34.05)/(546+ 273) = (1xxV_2)/273 or  V_2 = 11.35 ml

11.35 mL of vapour at 0^@C and 1 bar pressure weight = 0.0625 g

:. 22700 mL of vapour at 0^@C and 1 bar pressure will weigh

= 0.0625/11.35 xx 22700 = 125  g

:. Molar mass = 125 g mol^(-1)

Alternatively using

R  = 0.083 bar dm^3 K^(-1) mol^(-1)

PV = nRT, i.e n = (PV)/(RT) - (1.0 "bar" xx(34.05 xx 10^(-3) dm^(3)))/(0.003 "bar" dm^(3) K^(-1) Mol^(-1) xx 819 K)

= 5 xx 10^(-4) mol

:. Mass of 1 mole = 0.0625/(5 xx 10^(-4))g = 125 g

:. "Molar mass"  = 125 gmol^(-1)

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Solution 34.05 Ml of Phosphorus Vapour Weighs 0.0625 G at 546 °C and 0.1 Bar Pressure. What is the Molar Mass of Phosphorus? Concept: Ideal Gas Equation - Derivation of Ideal Gas Equation.
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