JEE MAIN - Chemistry (2016 - 9th April Morning Slot - No. 16)
The solubility of N2 in water at 300 K and 500 torr partial pressure is 0.01 g L−1. The solubility (in g L−1) at 750 torr partial pressure is :
0.0075
0.015
0.02
0.005
Explanation
Partial pressure = Mole fraction × Solubility
$${{{p_1}} \over {{p_2}}} = {{{s_1}} \over {{s_2}}}$$
$$ \Rightarrow $$ $${{500} \over {750}} = {{0.01} \over {{s_2}}}$$
$$ \Rightarrow $$ s2 = 0.015 g L-1
$${{{p_1}} \over {{p_2}}} = {{{s_1}} \over {{s_2}}}$$
$$ \Rightarrow $$ $${{500} \over {750}} = {{0.01} \over {{s_2}}}$$
$$ \Rightarrow $$ s2 = 0.015 g L-1
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