Question

When 2.25 g of sodium hydroxide (NaOH) was dissolved in 150.00 g of water a value of 11.00oC was obtained for ΔT.

1. Calculate the molarity of the sodium hydroxide solution.
2. Calculate the value (calories) for the heat of solution of 2.25 g of NaOH.
3. Calculate the number of calories that would be produced if one mole of sodium hydroxide was dissolved. (ΔHsolnNaOH)

Answer 1

Answer:

For 1: The molarity of sodium hydroxide solution is 0.375 M

For 2: The amount of heat absorbed by solution is 1674.75 Cal

For 3: The enthalpy change of the reaction when 1 mole of NaOH is dissolved is 1674.75 Cal

Explanation:

• For 1:

To calculate mass of a substance, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

Mass of water = 150.00 g

Putting values in above equation, we get:

$1g/mL=\frac{150.00g}{\text{Volume of water}}\\\\\text{Volume of water}=(1g/mL\times 150.00g)=150.00mL$

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Mass of solute (sodium hydroxide) = 2.25 g

Molar mass of sodium hydroxide = 40 g/mol

Volume of solution = 150.00 mL

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{2.25g\times 1000}{40g/mol\times 150mL}\\\\\text{Molarity of solution}=0.375M$

Hence, the molarity of sodium hydroxide solution is 0.375 M

• For 2:

To calculate the amount of heat absorbed, we use the equation:

$q=mc\Delta T$

where,

m = mass of solution = (2.25 + 150) = 152.25 g

c = specific heat capacity of water = 1 Cal/g.°C

$\Delta T$ = change in temperature = 11°C

Putting values in above equation, we get:

$q=152.25g\times 1Cal/g.^oC\times 11^oC\\\\q=1674.75Cal$

Hence, the amount of heat absorbed by solution is 1674.75 Cal

• For 3:

To calculate the enthalpy change of the reaction, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

$q$ = amount of heat absorbed = 16

74.75 Cal

n = number of moles = 1 mole

$\Delta H_{rxn}$ = enthalpy change of the reaction

Putting values in above equation, we get:

$\Delta H_{rxn}=\frac{1674.75Cal}{1mol}=1674.75Cal/mol$

Hence, the enthalpy change of the reaction when 1 mole of NaOH is dissolved is 1674.75 Cal