I 100 mL o a pharmaceutical preparation contains 20 mL o a 50% w/v solution o benzalkonium chloride, what is the percent strength o that agent in the solution

The percentage strength of the agent is 10 %

Since 100 mL of a pharmaceutical preparation contains 20 mL of a 50% w/v solution of benzalkonium chloride, we desire to find the percentage strength of that agent in the 100 mL solution.

Let C₁ = percentage strength of 20 mL solution = 50 %, V₁ = initial volume = 20 mL, C₂ = percentage strength of agent in 100 mL solution and V₂ = final volume = 100 mL.

Since the initial quantity of agent in 20 mL solution equals the final quantity of agent in 100 mL solution, we have that

C₁V₁ = C₂V₂

So, making C₂ = percentage strength of agent in 100 mL solution the subject of the formula, we have

C₂ = C₁V₁/V₂

Substituting the values of the variables into the equation, we have

C₂ = C₁V₁/V₂

C₂ = 50 % × 20 mL/100 mL

C₂ = 50 % × 0.2

C₂ = 10 %

So, the percentage strength of the agent is 10 %.

Learn more about percentage strength here:

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j