The room-temperature electrical conductivity of a semiconductor specimen is 2.8 × 10 4 (ω∙m)–1. if the electron concentration is 2.9 × 1022 m–3 and electron and hole mobilities are 0.14 and 0.023 m2/v∙s, respectively, calculate the hole concentration. (a) 1.24 × 1024 m–3 (b) 7.42 × 1024 m–3 (c) 7.60 × 1024 m–3 (d) 7.78 × 1024 m–3

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2