$(a)\;13.7\%\qquad(b)\;13\%\qquad(c)\;12\%\qquad(d)\;10\%$

$\large\frac{V_{mix}}{V_{Kr}} = \sqrt{\large\frac{M_{Kr}}{M_{mix}}}$

Or 1.16 = $\sqrt{(\large\frac{84}{M})}$

M = 62.425

For $Cl_2 \iff 2Cl$

$\large\frac{\text{Normal mol.wt}}{\text{Exp mol. wt.}} = 1 + \alpha$

$\large\frac{71}{62.425} = 1 + \alpha$

$\alpha = 0.137\; (or) \;13.7\%$

Hence answer is (A)

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