# What is the mass of a 405*mL block of aluminium given that rho_"Al"=2.70*g*mL^-1?

##### 2 Answers
Sep 11, 2017

Approx. $1.1 \cdot k g$

#### Explanation:

Now $\rho = \text{Mass"/"Volume}$, and thus....

$\text{Mass"=rho_"Al"xx"volume}$

=2.70*g*cancel(mL^-1)xx405*cancel(mL)=??*g

Sep 15, 2017

The mass of the Aluminum would be 1100 grams

#### Explanation:

$D = \frac{M}{V}$ so the density would be $2.7 \frac{g}{1} m l$

multiplying the Density by the volume will give the Mass

$\frac{M}{V} \times V = M$ putting the values into the equation gives

$2.7 \frac{g}{1} m l \times 405 m l = 1093.5 g$

However 2.7 has only two measured significant number so the answer can not be more precise than the least precise measurement so the answer can only have two significant digits.

Rounding 1093.5 down to two digits gives

1100 grams as the best answer.