What is the molality of a solution formed by dissolving 1.12 mol of kcl in 16.0 mol of water?

The molality of the solution is 3.45 mol/kg

Explanation:

Molality is a type of concentration which shows the moles of solute contained in  1 kg of solvent.

Moles of salt (KCl) = 1.12 moles

Moles of solvent (H₂O) = 18 moles

Let's convert the moles of solvent to mass (mol . molar mass)

18 m . 18g/m = 324 grams

We convert grams to kg → 324 g / 1000 = 0.324 kg

Now we can reach molality

1.12mol / 0.324 kg = 3.45 m

1 a

2 d

3 a

4 c

explanation:

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