Question

An Ostwald viscometer is calibrated using water at 20°C , . It takes 15.0 s for the fluid to fall from the upper to the lower level of the viscometer. A second liquid is then placed in the viscometer, and it takes 37.0 s for the fluid to fall between the levels. Finally, 100. mL of the second liquid weighs 76.5 g. What is the viscosity of the liquid?

Answer 1

Answer:

Viscosity of liquid $= 1.89$ cP

Explanation:

The viscosity of the liquid is determined by

$v = A* \rho* \delta t$

Where A represents the viscometer constant

ρ is the density of the fluid

∆t is the time required to  move through the viscometer.

First, we will derive the viscometer constant using the data for water  -

$v = A* \rho* \delta t$

$A = \frac{v}{\rho* \delta t}$

Substituting the given values in above equation, we get -

$\frac{1.0015* 10^{-2}* 0.1}{0.998*15*1000}$

$= 6.69 * 10^{-8}$ $m^2s^{-2}$

Viscosity of the second liquid is

$v = A* \rho* \delta t$

$v= 6.69 * 10^{-8} * 0.0765 * 37 * \frac{1}{1000} \\v = 0.0189 P\\v = 1.89 cP$

Answer 2

Answer:

The correct answer is 1.89 cP.

Explanation:

Viscosity can be calculated by using the equation:  

η = AρΔt, where A is the viscometer constant, η is the viscosity, Δt is the time, and ρ is density.  

The Δt given in the question is 15 s, the value of η is 1.0015 cP or 1.0015 * 10⁻² P, and the value of p is 0.998 g m/L

The value of viscometer constant can be determined by putting the values,  

A = η / ρΔt

A = (1.0015 * 10⁻² P / 0.998 kg L⁻¹ * 15 s) (0.1 kg/m/s / 1P) (1 m³/1000L)

= 6.69 * 10⁻⁶ * 10⁻² m² s⁻²

= 6.69 * 10⁻⁸ m² s⁻²

After finding the viscometer constant, the value of the second liquid viscosity can be calculated by putting the values,  

η = AρΔt

= 6.69 * 10⁻⁸ m² s⁻² (0.0765 kg / 0.1L) * 37s

= 189.36 * 10⁻⁸ m² s⁻¹ * (1000 L / 1 m³)

= 0.00189 kg m⁻¹ s⁻¹ (1 P / 0.1 Kg m⁻¹ s⁻¹)

= 0.0189 P

= 1.89 * 10⁻² P or 1.89 cP

Thus, the viscosity of the liquid will be 1.89 cP .