Could you please show me how to solve this problem ! Volume= 1.00cm^3,mass of water= 0.49 g, dry mass= 1.32 g, RD=2.70
To solve this problem, we need to find the relative density (RD) of a substance. Relative density is defined as the ratio of the density of a substance to the density of water. Here's how you can find the RD using the given information:
1. First, let's calculate the density of water using the mass and volume given. Density (ρ) is defined as mass (m) divided by volume (V). So, the density of water is:
ρ_water = m_water / V
Given: Volume = 1.00 cm^3 and mass of water = 0.49 g
ρ_water = 0.49 g / 1.00 cm^3
2. Now, let's calculate the density of the substance using the mass and volume given. The volume is the same as the volume of water (1.00 cm^3). The mass of the substance is given as the "dry mass."
Given: Volume = 1.00 cm^3 and dry mass = 1.32 g
ρ_substance = 1.32 g / 1.00 cm^3
3. Finally, we can calculate the relative density (RD) by dividing the density of the substance by the density of water:
RD = ρ_substance / ρ_water
Given: ρ_substance = 1.32 g / 1.00 cm^3 and ρ_water = 0.49 g / 1.00 cm^3
RD = (1.32 g / 1.00 cm^3) / (0.49 g / 1.00 cm^3)
RD = 2.70
Therefore, the relative density (RD) of the substance is 2.70.