Question

Which of the following has been arranged in order of decreasing freezing point?

• A. $0.05\text{M}KN{O}_3>0.04\text{M}CaC{l}_2>0.14\text{M}\text{sugar}>0.075\text{M}CuS{O}_4$
• B. $0.04\text{M}BaC{l}_2>0.14\text{M}\text{sucrose}>0.075\text{M}CuS{O}_4>0.05\text{M}KN{O}_3$
• C. $0.075MCuS{O}_4>0.14M\text{sucrose}>0.04MBaC{l}_2>0.05MKN{O}_3$
• D. $0.075\text{M}CuS{O}_4>0.05\text{M}NaN{O}_3>0.14\text{M}\text{sucrose}>0.04\text{M}BaC{l}_2$

Answer 1

The correct option is A $0.05\text{M}KN{O}_3>0.04\text{M}CaC{l}_2>0.14\text{M}\text{sugar}>0.075\text{M}CuS{O}_4$

Depression in freezing point is a colligative property and thus it depends on the numberof solute particles.

Number of moles of solute or ions in $1$ litre of the following solutions:
$0.05\text{M}KN{O}_3=2\times 0.05=0.1\text{mol}$
$0.04\text{M}CaC{l}_2=3\times 0.03=0.12\text{mol}$
$0.14\text{M}\text{sugar}=1\times 0.14=0.14\text{mol}$
$0.75\text{M}CuS{O}_4=2\times 0.75=0.15\text{mol}$
With increase in number of particles, freezing point decreases.
Hence, the order of freezing point is:
$0.05\text{M}KN{O}_3>0.04\text{M}CaC{l}_2>0.14\text{M}\text{sugar}>0.075\text{M}CuS{O}_4$