# the molality of pure gas A is 25/1.1 and pure gas B is 100/6.4. now a gaseous solution was prepared by mixing pure gas A and B. the molarity of gas A in gaseous solution at 10 atm and 200/0.821 K temperature is 0.1 M. find molality of gas a in the solution

2. So.assuming that equal volumes of gases A and B,say V are mixed in gaseous phase.

3. Calculation :

M = $\frac{{M}_{A}{V}_{A}+{M}_{B}{V}_{b}}{{V}_{A}+{V}_{B}}=\frac{{M}_{A}V+{M}_{B}V}{2V}\phantom{\rule{0ex}{0ex}}=\frac{{M}_{A}+{M}_{B}}{2}=\frac{\left({\displaystyle \frac{25}{1.1}}\right)+\left({\displaystyle \frac{100}{6.4}}\right)}{2}\phantom{\rule{0ex}{0ex}}=19.17M$

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