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A substance weight 5.74 g occupies a volume of `1.2cm^3.` Caluclate its density with due regard to significant digits. |
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Answer» Here, mass `(m)=5.74 g`, `volume(V)=1.2cm^(3)` As density. `rho=(mass)/(volume)=(5.74 g)/(1.2 cm^(3))=4.783 g cm^(3)` As mass has 3 significant digits and Volume has 2 significant digits therefore, as per rule, density will have only two significant digits, rounding off, we get `rho=4.8 cm^(-3)` |
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