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  • A 15 g Mixture of Hydrogen and helium is contained in a vessel at capacity. <img alt="0.225 \mathrm{dm}^3" src="https://entrancecorner.oncodecogs.com/gif.latex?0.225%20%5Cmathrm%7Bdm%7D%5E3" /

A 15 g Mixture of Hydrogen and helium is contained in a vessel at capacity. 0.225 \mathrm{dm}^3 at 5 bar and 27^{\circ} \mathrm{C}.
The Mass of Helium in the Mixture is ---g

Option: 1

14 g


Option: 2

18 g


Option: 3

16 g


Option: 4

12 g


Answers (1)

best_answer

\mathrm{P_V=\text { nmix } R J}

\mathrm{\begin{aligned} n_{\text {mis }}=\frac{5 \times 22.5}{.082 \times 300}=\frac{112-5}{24-6} & \approx 4 \text { mole } \end{aligned}}

out ot total 4 moles
Let Mole of He = x
Mole of \mathrm{H_2} =4-x
Total Mass of Mixture will be = \mathrm{=4 x+2(4-x)=15 g}

\mathrm{\text { Number of Mole of the gas } x=3.5 \text { mole }}

\mathrm{\text { Mars of the }=\text { Number of Mole } x \text { Molas Mass }}

\mathrm{\begin{aligned} & =3.5 \times 4 \\ & =14 \mathrm{~g} \end{aligned}}

Posted by

shivangi.shekhar

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