Question

18.72 g of a substance 'X' occupies 1.81 $c{m}^3$. What will be its density measured in correct significant figures?

• A. 10.3 g $c{m}^{-3}$
• B. 10.34 g $c{m}^{-3}$
• C. 10.4 g $c{m}^{-3}$
• D. 10.3425 g $c{m}^{-3}$

Answer 1

The correct option is A 10.3 g $c{m}^{-3}$

Density is mass per unit volume of the substance

$Density,\rho =\frac{mass\left(m\right)}{Volume\left(V\right)}$

Given :$m=18.72g$ and $V=1.81c{m}^3$

on substituting we get:

$\rho =\frac{18.72}{1.81}=10.3425$

Rule: Answer to a multiplication or division should be rounded off to the same number of significant figures as possessed by the least precise term in the calculation.

least precise is $1.81$ with 3 significant figures thus, the answer will also posses 3 significant figures as $10.3gc{m}^{-3}$

Hence, the correct option is $A$