Question

A. After the label fell off a bottle containing a clear liquid believed to be benzene, a chemist measured the density of the liquid to verify its identity. A 25.0-mL portion of the liquid had a mass of 21.95 gg. A chemistry handbook lists the density of benzene at 15∘C∘C as 0.878 g/mg/mL. Is the calculated density in agreement with the tabulated value?
B. An experiment requires 15.0 g of cyclohexane, whose density at 25oC is 0.7781 g/mL. What volume of cyclohexane should be used?
C. A spherical ball of lead has a diameter of 5.0 cm. What is the mass of the sphere if lead has a density of 11.34 g/cm3?

Answer 1

Answer:

A. Yes, the calculated density in agreement with the tabulated value.

B. 19.28 mL of volume of cyclohexane should be used.

C. 742.20 is the mass of the sphere of lead.

Explanation:

A.

Volume of the liquid = V = 25.0 mL

Mass of the liquid = m = 21.95 g

Density of the liquid = d

$d=\frac{m}{V}$

$=\frac{21.95 g}{25.0 mL}=0.878 g/mL$

Density mentioned in the report book = d' = 0.878 g/mL

d' = d = 0.878 g/mL

Yes, the calculated density in agreement with the tabulated value.

B.

Volume of the liquid cyclohexane= V = ?

Mass of the liquid cyclohexane= m = 15.0 g

Density of the liquid cyclohexane = d = 0.7781 g/mL

$d=\frac{m}{V}$

$V=\frac{15.0 g}{0.7781 g/mL}=19.28 mL$

19.28 mL of volume of cyclohexane should be used.

C.

Diameter of the ball = d = 5.0 cm

Radius of the ball = r = 0.5 × d = 2.5 cm

Volume of sphere ,V= $\frac{4}{3}\pi r^3$

$V = \frac{4}{3}\times 3.14\times (0.25 cm)^3=65.45 cm^3$

Volume of the spherical lead ball = V  

Mass of the  spherical lead ball= m = ?

Density of the  spherical lead ball = d = $11.34 g/cm^3$

$d=\frac{m}{V}$

$m=d\times V=11.34 g/cm^3\times 65.45 cm^3=742.20 g$

742.20 is the mass of the sphere of lead.