Question

Solid polymeric particles of diameter 20 \(\mu\)m are suspended in water such that the concentration of polymeric particles in the solution is 1% weight/volume (w/v). If the mass density of the polymeric particle is \(1 \text{ g/cm}^3\), the number of polymeric particles present in 1 mL of this solution is \(\times 10^6\).
(Round off to one decimal place)

Hint:

Use "1% w/v" to get the mass of particles per mL, convert that mass to a volume using density (1 g/cm^3), then divide by the volume of a single spherical particle found from its diameter.

Answer 1

Correct Answer: 2.4

Step 1: Understand the concentration.
"1% weight/volume (w/v)" means 1 gram of solute (the polymeric particles) is present in every 100 mL of solution.
So the mass of particles in 1 mL of solution is
\[ m = \frac{1 \text{ g}}{100 \text{ mL}} \times 1 \text{ mL} = 0.01 \text{ g} \]

Step 2: Convert this mass to the total volume of solid particle material.
The mass density of the polymeric particle is \(\rho = 1 \text{ g/cm}^3\), and since 1 mL \(= 1 \text{ cm}^3\), density directly converts mass to volume:
\[ V_{\text{total particles}} = \frac{m}{\rho} = \frac{0.01 \text{ g}}{1 \text{ g/cm}^3} = 0.01 \text{ cm}^3 \]
This is the combined solid volume of every particle present in 1 mL of the suspension.

Step 3: Find the volume of one single particle.
Diameter \(d = 20\ \mu\text{m}\), so radius \(r = 10\ \mu\text{m}\). Convert to cm using \(1\ \mu\text{m} = 10^{-4}\) cm:
\[ r = 10 \times 10^{-4} \text{ cm} = 1\times 10^{-3} \text{ cm} \]
Treat each particle as a sphere, so its volume is
\[ v = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (1\times 10^{-3})^3 \text{ cm}^3 = \frac{4}{3}\pi \times 1\times 10^{-9} \approx 4.19\times 10^{-9} \text{ cm}^3 \]

Step 4: Divide the total particle volume by the volume of one particle.
Number of particles \(N\) present in 1 mL:
\[ N = \frac{V_{\text{total particles}}}{v} = \frac{0.01}{4.19\times 10^{-9}} \approx 2.39\times 10^{6} \]
Rounded to one decimal place (in units of \(\times 10^6\)), \(N \approx 2.4\times 10^6\).

Final Answer:
The number of polymeric particles present in 1 mL of the solution is about \(2.4\times 10^6\) (the accepted official range for this question is 2.0 to 3.0).
\[ \boxed{N \approx 2.4\times 10^6} \]