calculate the density in gram per ml of 0.6 litre of solution weighing 180 gram
Answers
Answered by
1
Explanation:
As we know that the density of the solution is given by the formula
d = \frac{m}{v}d=
v
m
m= 180g
\begin{gathered} m = {180g} \\ \\ v = 0.6 : l > 0.6 \times 1000 > 600ml \\ on \: putting \: the \: values \: in \: the \: formula \\ \\ d = \frac{180}{600} > 0.3g {ml}^{ - 1} \end{gathered}
m=180g
v=0.6:l>0.6×1000>600ml
onputtingthevaluesintheformula
d=
600
180
>0.3gml
−1
Answered by
1
The density of water is approximately 1g/ml.
Thus, 1g of water occupy volume = 1 ml
The molar mass of water = 18 g/mol
Number of moles of water in 1g =
18
1
moles = 0.055 moles H
2
O
Now 18 g water contain = 6.022 × 10
23
moleculesH
2
O
Thus 1 g water will have =
18
6.022×10
23
molecules H
2
O = 0.334 × 10
23
molecules H
2
O
This shows that
0.334 ×10
23
molecules H
2
O will occupy volume = 1 ml
Therefore volume occupied by 1 H
2
O molecule =
0.334×10
23
1
=2.99 × 10
−23
ml
Thus volume occupied by 1 H
2
O mlecule is 2.99 × 10
−23
ml.
Thus, 1g of water occupy volume = 1 ml
The molar mass of water = 18 g/mol
Number of moles of water in 1g =
18
1
moles = 0.055 moles H
2
O
Now 18 g water contain = 6.022 × 10
23
moleculesH
2
O
Thus 1 g water will have =
18
6.022×10
23
molecules H
2
O = 0.334 × 10
23
molecules H
2
O
This shows that
0.334 ×10
23
molecules H
2
O will occupy volume = 1 ml
Therefore volume occupied by 1 H
2
O molecule =
0.334×10
23
1
=2.99 × 10
−23
ml
Thus volume occupied by 1 H
2
O mlecule is 2.99 × 10
−23
ml.
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