The mass density of pure water is 1,000 kg/m³. If the molecular mass of water is 18.0 u, how many water molecules are in spherical water balloon of radius 1.0 cm? A) 1.0 × 10²³ B) 2.5 × 10²⁴ C) 1.4 × 10²³ D) 4.2 × 10²³

Question

Quizplus · Accepted Answer

The volume of the spherical water balloon can be calculated using the formula: 
V = (4/3)πr3 
where r = 1 cm 
Therefore, V = (4/3)π(1 cm)3 = 4.19 cm3 
Now, we can calculate the mass of the water in the balloon using the density formula: 
m = ρV 
where ρ is the density of water = 1000 kg/m3, and V is the volume calculated above in cm3. 
Converting the volume to m3: 
V = 4.19 × 10-6 m3 
m = 1000 kg/m3 × 4.19 × 10-6 m3 = 0.00419 kg 
Next, we can calculate the number of water molecules using the Avogadro's number: 
n = m/M 
where M is the molar mass of water = 18 g/mol = 0.018 kg/mol 
n = 0.00419 kg / 0.018 kg/mol = 0.232 mol 
Number of water molecules = n × N_A 
where N_A is Avogadro's number = 6.022 × 1023 mol-1 
Number of water molecules = 0.232 mol × 6.022 × 1023 mol-1 = 1.397 × 1023 
Therefore, the best answer is C, 1.4 × 1023.

The Mass Density of Pure Water Is 1,000 Kg/m3

The Mass Density of Pure Water Is 1,000 Kg/m³