A certain substance X has a normal freezing point of −3.1°C and a molal freezing point depression constant =Kf·6.23°C·kgmol−1. Calculate the freezing point of a solution made of 12.3g of urea NH22CO dissolved in 300.g of X. Be sure your answer has the correct number of significant digits.

Freezing T° of solution =  - 7.35 °C

Explanation:

This is about the freezing point depression, a colligative property which depends on solute.

The formula is: Freezing T° pure solvent - Freezing T° solution = m . Kf . i

Freezing T° of pure solvent is -3.1°C

At this case, i = 1. As an organic compound the urea does not ionize.

We determine the molality (mol/kg of solvent)

We convert mass to moles:

12.3 g . 1mol / 60.06 g = 0.205 moles

0.205 mol / 0.3 kg = 0.682 mol/kg

We replace data in the formula:

-3.1°C - Freezing T° of solution  = 0.682 mol/kg . 6.23 kg°C/mol . 1

Freezing T° of solution = 0.682 mol/kg . 6.23 kg°C/mol . 1 + 3.1°C

Freezing T° of solution =  - 7.35 °C

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