Question

an empty density bottle weighed 23.5 and 48.4g when filled with water .The empty bottle was partially filled with sand until and it weighed 36.5g water was added to the bottle of sand until it was completely filled it then weighed 56.5g determine the density of the sand, ( density of water = 1g/cm3)​

Answer 1

Answer:

Density of Sand is $2.653g/cm^{3}$.

Explanation:

Given Empty Density bottle weighs 23.5gm(W=23.5gm)

Weight of bottle when completely filled water=48.4gm

So amount of water required to fill the bottle=Weight of bottle filled with water-W

Amount of water required to fill the bottle($w_{max}$)=48.4gm-23.5gm

$w_{max}=24.9$g

Since we know density of water $d_{w} =1g/cm^{3}$ and $w_{max}$

We can calculate volume of empty space in the bottle(V).

$w_{max}=d_{w}$$\times$V

V=$\frac{w_{max} }{d_{w} }$

V=$\frac{24.9}{1}$

V=24.9 $g/cm^{3}$

Now bottle is partially filled with sand,and weight of bottle is ($w_{s}$)36.5gm

So,

Amount of sand added ($m_{s}$)=36.5-Weight of the bottle

$m_{s}$=13g

After filling the bottle with water again,the weight of the bottle becomes ($W_{2}$=56.5g)

Therefore,

amount of water added to the bottle of sand in grams = $W_{2}$-36.5gm

amount of water added =56.5g-36.5g

amount of water added =20g

As the density of water = 1g/$cm^{3}$

Amount of water (in grams )=Volume of water occupied

20=volume of water added

Therfore volume of water added to the sand filled bottle($V_{w}$)=20$cm^{3}$

As we know the total volume of the water bottle(V),

Volume of the sand occupied in the water bottle=V-$V_{w}$

$V_{s}$=24.9g-20g

$V_{s}$=4.9g

We know,

Density=Mass/Volume

Therefore,

density of sand = $\frac{m_{s} }{V_{s} }$

density of sand =$\frac{13}{4.9}$

density of sand = $2.653g/cm^{3}$