Achemist dissolves 1 mol nacl in a 2.0-kg sample of water, then dissolves 1 mol sucrose in another 2.0-kg sample of water. why does the nacl solution have the lower freezing point?

Due to more number of solute particles in NaCl, it will have lower freezing point.

Explanation:

Lowering in freezing point is a colligative property. It depends upon the number of solute particles.

Now when we dissolve NaCl and sucrose in equal moles in two same mass of samples of water, the number of particles in NaCl solution is more as compared to sucrose.

NaCl being an electrolyte will dissociate into its ions (Na⁺ and Cl⁻). Thus each mole of NaCl will give more than one mole of solute particles.

Sucrose being a non electrolyte, will not dissociate and thus its one mole will give only one mole of solute particles.

Depression in freezing point of NaCl will be thus more than that of sucrose solution.

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol