A drop (0.05 mL) of a solution contains 3.0× 10- 6 moles of H+ Ions. If the rate constant of disappearance of H+ ion is 1.0× 107 mol litre- 1sec- 1, the time taken for H+ ions in the drop to disappear is p× 10- q. Calculate the value of q-p . Give the nearest integer as the answer.

Question

A drop (0.05 mL) of a solution contains 3.0× 10- 6 moles of H+ Ions. If the rate constant of disappearance of H+ ion is 1.0× 107 mol litre- 1sec- 1, the time taken for H+ ions in the drop to disappear is p× 10- q. Calculate the value of q-p . Give the nearest integer as the answer. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Concentration of drop =( text mole / text volume in mL ) × 1000 =(3 × 10-6/0.05) × 1000=0.06 mole litre -1 Rate of disappearence =( text conc.change / text time ) 1 × 107= (0.06/ text time ) Time =6 × 10-9 Sec So, q-p=3

Q. A drop (0.05mL) of a solution contains 3.0×10−6 moles of H+ Ions. If the rate constant of disappearance of H+ ion is 1.0×107mollitre−1sec−1, the time taken for H+ ions in the drop to disappear is p×10−q. Calculate the value of q−p . Give the nearest integer as the answer.

Concentration of drop = volume in mL mole ​×1000 =0.053×10−6​×1000=0.06 mole litre −1 Rate of disappearence = time conc.change ​1×107= time 0.06​ Time =6×10−9 Sec So, q−p=3

Concentration of drop $= \frac{mole}{volume in mL} \times 1000$
$= \frac{3 \times 1 0 ^{- 6}}{0.05} \times 1000 = 0.06$ mole litre $^{- 1}$
Rate of disappearence $= \frac{conc.change}{time} 1 \times 1 0^{7} = \frac{0.06}{time}$
Time $= 6 \times 1 0^{- 9}$ Sec So, $q - p = 3$