Question

Accounting for buoyancy, what apparent mass of CsCl (density = 3.988g/mL) in air should you weight out to obtain a true mass of 1.267g?

Short answer

The required true mass is 1.266 g.

Step-by-step solution

Defining density

Density is calculated using the formula d = M/V , where d represents density, M is mass, and V is volume.

Explaining the formula

The formula required to solve this problem is

$m=\frac{{\displaystyle \stackrel{,}{m}}\left(1-{\displaystyle \frac{d_a}{d_w}}\right)}{\left(1-\frac{d_a}{d}\right)}$.

Here,

m is the mass read on the balance in units of (g) ,

$d_a$is the density of air, which is 0.0012g/mL near 1 bar,

$d_w$is the density of the calibration weights, which is 8.0 g/mL ,

d is the density of the object being weighed in units of (g/mL) , and

m is the mass of the object weighed in vacuum in units of (g) .

Finding the mass of CsCl weighed in air

Buoyancy corrections are weight measurements that are adjusted by estimating the difference in densities between the standard and unknown weights.

We will apply the formula for buoyancy adjustment given below.

Here, stands for the density of air, whereas stands for the density of the weighed product in the air.

$m=\frac{m\left(1-{\displaystyle \frac{d_a}{d_w}}\right)}{\left(1-\frac{d_a}{d_o}\right)}$.

In order to find the mass of CsCl (density = 3.988 g/mL) weighed in air, we will use the true mass of CsCl , which is 1.267 g .

$\begin{array}{l}1.267g=\stackrel{,}{m}+\frac{(1-*(0.0012g/mL/8g/mL)}{1-(0.0012g/mL/3.988g/mL}\\ 1.267g=\stackrel{,}{m}+1.00051\\ m=1.2668g\end{array}$.