Three different containers hold three different gases with the following parameters: Container 1: 67.2 L containing gas X at 0°C and 1.0325 bar. Container 2: 44.8 L containing gas Y at 546 K and 760 m

Question

Three different containers hold three different gases with the following parameters:

Container 1: 67.2 L containing gas X at 0°C and 1.0325 bar.

Container 2: 44.8 L containing gas Y at 546 K and 760 mm of Hg.

Container 3: 89.6 L containing gas Z at 101325 N/m² and 0°C.

These gases are then mixed in a single container. If the gram molecular masses of X, Y, and Z are 10 g/mol, 20 g/mol, and 40 g/mol, respectively,

which of the following statements are correct?

TestHub · Accepted Answer

To solve this problem, we first calculate the moles of each gas using the ideal gas law, $P V = n RT$ .

Given constants:

$R = 0.0821 L a t mm o l^{- 1} K^{- 1}$

$R = 8.314 J m o l^{- 1} K^{- 1}$

$1 a t m = 1.01325 ba r = 760 mm H g = 101325 N / m^{2} (P a)$

$0^{0} C = 273.15 K$

Molar mass of Ca = $40 g / m o l$

Avogadro's number, $N_{A} = 6.022 \times 1 0^{23} m o l^{- 1}$

Container 1 (Gas X):

$V_{X} = 67.2 L$

$T_{X} = 0^{0} C = 273.15 K$

$P_{X} = 1.0325 ba r = 1.0325 ba r \times \frac{1 a t m}{1.01325 ba r} = 1.019 a t m$

$n_{X} = \frac{P _{X} V _{X}}{R T _{X}} = \frac{1.019 a t m \times 67.2 L}{0.0821 L a t mm o l ^{- 1} K ^{- 1} \times 273.15 K} = 3.04 m o l \approx 3 m o l$

Container 2 (Gas Y):

$V_{Y} = 44.8 L$

$T_{Y} = 546 K$

$P_{Y} = 760 mm H g = 1 a t m$

$n_{Y} = \frac{P _{Y} V _{Y}}{R T _{Y}} = \frac{1 a t m \times 44.8 L}{0.0821 L a t mm o l ^{- 1} K ^{- 1} \times 546 K} = 0.999 m o l \approx 1 m o l$

Container 3 (Gas Z):

$V_{Z} = 89.6 L$

$T_{Z} = 0^{0} C = 273.15 K$

$P_{Z} = 101325 N / m^{2} = 1 a t m$

$n_{Z} = \frac{P _{Z} V _{Z}}{R T _{Z}} = \frac{1 a t m \times 89.6 L}{0.0821 L a t mm o l ^{- 1} K ^{- 1} \times 273.15 K} = 3.99 m o l \approx 4 m o l$

Total moles in the mixture: $n_{t o t a l} = n_{X} + n_{Y} + n_{Z} = 3 + 1 + 4 = 8 m o l$

A. Mole percent of gas Z:

Mole percent of Z $= \frac{n _{Z}}{n _{t o t a l}} \times 100% = \frac{4 m o l}{8 m o l} \times 100% = 50%$

Statement A is correct.

B. Average molecular mass of final mixture:

Average molecular mass $= \frac{( n _{X} \times M _{X} ) + ( n _{Y} \times M _{Y} ) + ( n _{Z} \times M _{Z} )}{n _{t o t a l}}$

Average molecular mass $= \frac{( 3 m o l \times 10 g / m o l ) + ( 1 m o l \times 20 g / m o l ) + ( 4 m o l \times 40 g / m o l )}{8 m o l}$

Average molecular mass $= \frac{30 + 20 + 160}{8} = \frac{210}{8} = 26.25 g / m o l$ (or amu)

Statement B is correct.

C. Mole ratio of gas X and gas Z:

Mole ratio of X to Z $= \frac{n _{X}}{n _{Z}} = \frac{3 m o l}{4 m o l} = \frac{3}{4}$

Statement C is correct.

D. Number of molecules present in final mixture is equal to number of atoms in 320 g of Ca:

Number of molecules in final mixture $= n_{t o t a l} \times N_{A} = 8 m o l \times N_{A}$

Number of atoms in 320 g of Ca $= \frac{ma sso f C a}{m o l a r ma sso f C a} \times N_{A} = \frac{320 g}{40 g / m o l} \times N_{A} = 8 m o l \times N_{A}$

Since both values are $8 N_{A}$ , statement D is correct.

All statements A, B, C, and D are correct.

Chemistry - Ideal Gas Question with Solution | TestHub

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JEE Advanced 2028 (Class 11)