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## Q. 5.1

A 10-mL volumetric pipet was calibrated following the procedure just outlined, using a balance calibrated with brass weights having a density of 8.40 g/cm³. At 25 °C the pipet was found to dispense 9.9736 g of water. What is the actual volume dispensed by the pipet?

## Verified Solution

At 25 °C the density of water is 0.99705 g/cm³. The water’s true weight, therefore, is

$W_{\mathrm{v}}=9.9736 g\times \left[1+\left\lgroup\frac{1}{0.99705}-\frac{1}{8.40}\times 0.0012 \right\rgroup\right] = 9.9842 g$

and the actual volume of water dispensed by the pipet is

$\frac{9.9842 g}{0.99705 g/cm^3} =10.014 cm^3 =10.014 ml$

If the buoyancy correction is ignored, the pipet’s volume is reported as

$\frac{9.9736 g}{0.99705 g/cm^3} =10.003 cm^3 =10.003 ml$

introducing a negative determinate error of –0.11%.