Please elaborate with your answers .Thanks ! 3. The density of a liquid is to be
ID: 1024540 • Letter: P
Question
Please elaborate with your answers .Thanks !
3. The density of a liquid is to be determined by measuring the mass of the liquid delivered by a pipet. The mass of the weighing bottle and stopper is 19.9314 t 0.0005 g. 10 t 0.02 mL of the liquid is delivered to the weighing bot- tle, stoppered, and weighed yielding a mass of 28.1038 t0.0005 g. What is the density of the liquid and its associ ated error? 4. The volume of a cylindrical capillary tube is given by v mr h where r is its radius and h its height. If the radius of the capillary tube is found to be 0.030 t0.002 cm and the height found to be 4.0 t 0.1 cm, what is the vol- ume of the capillary tube and its associated error? What measurement must be made to a higher precision to decrease the error in the volumeExplanation / Answer
3. The mass of the weighing bottle before addition of the liquid is 19.9314 ± 0.0005 gm; the mass of the bottle after addition of the liquid is 28.1038 ± 0.0005 gm. Therefore, the weight of the liquid added = (28.1038 - 19.9314) gm = 8.1724 gm.
The volume of liquid transferred = 10 ± 0.02 mL. The density of the liquid is (weight of liquid transferred)/(volume of liquid transferred) = 8.1724 gm/10 mL = 0.81724 gm/mL.
Relative uncertainty in the weight of the liquid transferred = 0.0005 gm/8.1724 gm = 6.1181*10-5.
Relative uncertainty in the volume of the liquid = 0.02 mL/10 mL = 2.*10-3
Therefore, relative uncertainty in the density of the liquid = [(6.1181*10-5)2 + (2.0*10-3)2]1/2 = (4.004*10-6)1/2 = 2.0009*10-3 2.0*10-3.
Absolute uncertainty in the density of the liquid = (relative uncertainty in density)*(density calculated) = (2.0*10-3)*(0.81724 gm/mL) = 1.6345*10-3 gm/mL 0.0016 gm/mL.
Therefore, the calculated density of the liquid is 0.81724 ± 0.0016 gm/mL (ans).
4. The volume of the capillary is given by V = r2h where r = 0.030 ± 0.002 cm and h = 4.0 ± 0.1 cm. The volume is V = (3.14)(0.030 cm)2(4.0 cm) = 0.011304 cm3 1.1304*10-2 cm3.
Uncertainty in radius = 0.002 cm; relative uncertainty in the radius is 2*(0.002 cm/0.030 cm) = 0.1333 (the factor 2 is added since the radius appears as a square).
Uncertainty in height = 0.1 cm; relative uncertainty in height = 0.1 cm/4.0 cm = 0.025.
Relative uncertainty in volume = [(0.1333)2 + (0.025)2]1/2 = 0.1356; absolute uncertainty in volume = (0.1356)*(1.1304*10-2 cm3) = 1.5328*10-3 cm3 1.5*10-3 cm3. The volume of the cylinder is 0.011304 ± 0.0015 cm3 (ans).
Since the radius appears as square in the volume expression, accurate determination of the radius is more important since a small error in the measurement of the radius will increase the error in the volume by two-folds (ans).