JEE MAIN - Chemistry (2020 - 3rd September Evening Slot - No. 19)
If 250 cm3
of an aqueous solution containing 0.73 g of a protein A is isotonic with one litre of
another aqueous solution containing 1.65 g of a protein B, at 298 K, the ratio of the molecular
masses of A and B is ______ × 10–2 (to the nearest integer).
Answer
177
Explanation
Let molar mass of protein A = x g/mol
Let molar mass of protein B = y g/mol
$$\pi $$A = osmotic pressure of protein A = $${{0.73} \over x} \times {{1000} \over {250}} \times RT$$
$$\pi $$B = osmotic pressure of protein B = $${{1.65} \over y} \times {1 \over 1} \times RT$$
Given $$\pi $$A = $$\pi $$B
$$ \Rightarrow $$ $${{0.73} \over x} \times {{1000} \over {250}} \times RT$$ = $${{1.65} \over y} \times {1 \over 1} \times RT$$
$$ \Rightarrow $$ $${x \over y}$$ = $${{0.73} \over {0.25 \times 1.65}}$$ = 1.77
= 177 $$ \times $$ 10-2
Let molar mass of protein B = y g/mol
$$\pi $$A = osmotic pressure of protein A = $${{0.73} \over x} \times {{1000} \over {250}} \times RT$$
$$\pi $$B = osmotic pressure of protein B = $${{1.65} \over y} \times {1 \over 1} \times RT$$
Given $$\pi $$A = $$\pi $$B
$$ \Rightarrow $$ $${{0.73} \over x} \times {{1000} \over {250}} \times RT$$ = $${{1.65} \over y} \times {1 \over 1} \times RT$$
$$ \Rightarrow $$ $${x \over y}$$ = $${{0.73} \over {0.25 \times 1.65}}$$ = 1.77
= 177 $$ \times $$ 10-2
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