Question

The surface tension of water was determuned in a laboratory by using the drop weight method. 100 drops were released from a burette the inner diameter of whose opening is 1.8mm. The mass of the droplets was 3.78g dertermine the surface tension of the water and comparing it with the tabulatef value

Answer 1

Answer:

The surface tension of the water is 6.278×10⁻² N/m

error = 13.65%

Explanation:

The surface tension of water is given by

$\gamma = \frac{F}{L}$

Where F is the force acting on water and L is the length over which is force is acted.

We are given the mass of 100 droplets of water

M = 3.78 g

n = 100

The mass of 1 droplet is given by

$m = \frac{M}{n} \\\\m = \frac{3.78}{100}\\\\m = 0.0378 \: g \\\\m = 3.780\times10^{-5} \: kg$

The force acting on a single droplet of water is given by

$F = m \cdot g$

Where m is the mass of water droplet and g is the acceleration due to gravity

$F = 3.780\times10^{-5} \cdot 9.81$

$F = 3.708\times10^{-4} \: N$

The circumferential length of the droplet is given by

$L = \pi \cdot d$

Where d is the diameter

$L = \pi \cdot 1.88\times10^{-3}\\\\L = 5.906 \times10^{-3} \: m$

Now we can find out the required surface tension of the water

$\gamma = \frac{3.708\times10^{-4} }{5.906 \times10^{-3}} \\\\\gamma = 0.06278\: N/m\\\\\gamma = 6.278 \times10^{-2} \: N/m$

Therefore, the surface tension of the water is 6.278×10⁻² N/m

The tabulated value of the surface tension of water at 20 °C is given by

$\gamma_t = 0.0727 \: N/m$

The percentage error between tabulated and calculated surface tension is given by

$error = \frac{\gamma_t - \gamma }{\gamma_t}$

$error = \frac{ 0.0727 - 0.06278}{0.0727} \times 100\%$

$error = 13.65 \%$