A substance of mass 49.53 g occupies 1.5 cm | vedclass

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(B) Density is defined as the ratio of mass to volume: $	ext{Density} = \frac{	ext{Mass}}{	ext{Volume}}$.Given: $	ext{Mass} = 49.53 \ g$ (which ha

Vedclass · Accepted Answer

$33$

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(B) Density is defined as the ratio of mass to volume: $	ext{Density} = \frac{	ext{Mass}}{	ext{Volume}}$.Given: $	ext{Mass} = 49.53 \ g$ (which has $4$ significant figures).Given: $	ext{Volume} = 1.5 \ cm^3$ (which has $2$ significant figures).Calculation: $	ext{Density} = \frac{49.53}{1.5} = 33.02 \ g \ cm^{-3}$.According to the rules of significant figures,when dividing,the result should have the same number of significant figures as the measurement with the fewest significant figures.Since $1.5$ has $2$ significant figures,the final answer must be rounded to $2$ significant figures.Therefore,the density is $33 \ g \ cm^{-3}$.

$A$ substance of mass $49.53 \ g$ occupies $1.5 \ cm^3$ of volume. The density of the substance (in $g \ cm^{-3}$) with the correct number of significant figures is:

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