What is the final temperature of the solution formed when 1.52 g of NaOH is added to 35.5 g of water at 20.1 [infinity]C in a calorimeter?

NaOH (s) ---> Na +(aq) + OH –(aq) DH = -44.5 kJ/mol

THE FINAL TEMPERATURE OF THE SOLUTION IS 31.4°C.

Explanation:

To solve for the final temperature of the solution formed, we must first calculate the quantity of heat given off by the reaction.

From the equation of the reaction, it will be observed that ΔH is negative meaning that the heat is given off from the reaction.

Quantity of heat = ΔH * number of mole of NaOH

So, we calculate the number of moles of NaOH used by dividing the mass of NaOH by the relative molecular mass.

RMM of NaOH = 40 g/mol

Number of mole = mass / RMM

n = 1.52 / 40 = 0.038 moles.

So therefore, Q =ΔH * n (NaOH)

Q = 44.5 * 0.038

Q = 1.691 kJ

We can now solve for the final temperature by using the formula assuming that the specific heat of water is 4 2 J/g °C

Q = mc (T2 - T1)

Rearranging,

T2 = Q / m c + T1

T2 = 1.691 kJ / 35.5 g * 4.2 J/g°C + 20.1° C

T2 = 1.691* 10^3 J / 149.1 J ° C + 20.1 °C

T2 = 0.0113 * 10^3 + 20.1

T2 = 11.3 + 20.1 ° C

T2 = 31.4 °C

The final temperature of the solution formed is 31.4 °C.

the answer is b. dilution