Anon-ideal solution was prepared by mixing 30 mL chloroform and 50 mL acetone. The volume of mixture will be :

Question

Anon-ideal solution was prepared by mixing 30 mL chloroform and 50 mL acetone. The volume of mixture will be : (A) > 80 mL (B) < 80 mL (C) = 80 mL (D)  ge 80 mL

Tardigrade Admin · Accepted Answer

Chloroform and acetone form a non-ideal solution in which A...B type interaction are more than A.. A and B...B type interaction due to H-bonding. Hence, the solution shows, negative deviation from Raoult's law, i.e.,

Δ V_{mi x} = - v e; Δ H_{mi x} = - v e

∴

total volume of solution = less than (30 + 50 mL) or < 80 mL

Q. Anon-ideal solution was prepared by mixing $30 m L$ chloroform and $50 m L$ acetone. The volume of mixture will be :

> 80 mL

< 80 mL

= 80 mL

$\geq$ 80 mL

Q. Anon-ideal solution was prepared by mixing 30mL chloroform and 50mL acetone. The volume of mixture will be :

> 80 mL

< 80 mL

= 80 mL

≥ 80 mL

Q. Anon-ideal solution was prepared by mixing $30 m L$ chloroform and $50 m L$ acetone. The volume of mixture will be :

$\geq$ 80 mL