A substance weight 5.74 g occupies a volume of `1.2cm^3.` Caluclate its density with due regard to significant digits.

Options

A. `4.8 cm^(-3)`
B. `4.2 cm^(-3)`
C. `4.1 cm^(-3)`
D. `4.5 cm^(-3)`

Answer

A

Explanation

Here, mass `(m)=5.74 g,` volume `(v)=1.2cm^(3)`
As density, `P=("mass")/("volume")=(5.74g)/(1.2cm^(3))=4.783gcm^(3)`
As mass has 3 significant digits and volume has only two significant digits, rounding off, we get `p=4.8cm^(-3)`.