Question

A)assuming volumes are additive, what volume of water must be added to 28.0 mL of 1.25 M FeCl3 to prepare a 0.650 M FeCl3 solution?
B) what mass of Ca(OH)2 would be required to neutralize 120.0 mL of 0.350 M HClO4? molar mass= Ca(OH)2 74.08; HClO4 100.46
C) what is the minimum volume of 0.08475 M KOH needed to neutralize 10.00 mL of 0.1801 M H2C2O4?
D) a 10.0 L sample of an ideal gas at 25.0°C is heated at constant pressure until the volume has doubled. what is the final temp?

Answer 1

A) dilution formula is M1V1= M2V2

M1 = 1.25 M, M2 = 0.65 M,

V1 = 28 mL, V2 = ?

V2 = M1V1/M2 = (28 . 1.25)/0.65

V2 = 53.85 mL needed to add.

B) 2 HClo₄ + Ca(OH)₂ = Ca(Clo₄)₂ + 2 H₂O

here, one mole of ca(OH)2 requied to quench two moles of HClO4

no.of moles of HClO4 = molarity mltiply with volume in lit.

Given that, moles of HClO4 = 0.35 . 0.12 = 0.042 moles

so, 0.21 moles of Ca(OH)2 is requied to neutralize 0.42 moles of HClO4.

mass of Ca(OH)2 is equals to 0.21 multiply by74.08.

i.e., 15.56 g of Ca(OH)2 needed.

C) H2C2O4.2H2O + 2KOH = K2C2O4 + 4H2O

no.of moles of H2C2O4 = 0.1801 multiply by 0.01 lt. = 0.001801 moles

two moles of KOH is required per each mole of K2C2O4.

i.e., volume = 0.08475/0.003602 = 25.52 lt.

D) according to charles law, If the pressure of a gas sample is kept constant, the volume of the sample will vary directly with the temperature in Kelvin (Figure 9.9). As the temperature increases, so will the volume; if the temperature decreases, the volume will decrease. This relationship can be expressed by an equation relating the initial volume (V₁) and initial temperature (T₁ measured in K) to the final volume (V₂) and final temperature (T₂ measured in K). At constant pressure,

V1V2

=

T1T2

V1 = 10 L, V2 = 20 L, T1 = 298 K, T2 = ?

T2 = 2(298) = 596 K