JEE MAIN - Chemistry (2019 - 9th April Morning Slot - No. 7)
The osmotic pressure of a dilute solution of an
ionic compound XY in water is four times that
of a solution of 0.01 M BaCl2 in water.
Assuming complete dissociation of the given
ionic compounds in water, the concentration of
XY (in mol L–1) in solution is :
4 × 10–4
6 × 10–2
4 × 10–2
16 × 10–4
Explanation
Given that, osmotic pressure of XY($${\pi _{XY}}$$) in water is four times that
of a solution of 0.01 M BaCl2($${\pi _{BaC{l_2}}}$$).
$$ \therefore $$ $${\pi _{XY}}$$ = 4 $$ \times $$ $${\pi _{BaC{l_2}}}$$
$$ \Rightarrow $$ 2 [XY] = 4 $$ \times $$ 3 $$ \times $$ 0.01
$$ \Rightarrow $$ [XY] = 0.06 = 6 × 10–2 mol L–1
$$ \therefore $$ $${\pi _{XY}}$$ = 4 $$ \times $$ $${\pi _{BaC{l_2}}}$$
$$ \Rightarrow $$ 2 [XY] = 4 $$ \times $$ 3 $$ \times $$ 0.01
$$ \Rightarrow $$ [XY] = 0.06 = 6 × 10–2 mol L–1
Comments (0)
