Answer:
B. 0.2.
Explanation:
<em>n = mass/molar mass</em>
mass of CaCO₃ = 20 g, molar mass of CaCO₃ = 100.0869 g/mol.
<em>∴ n = mass/molar mass = </em>(20 g)/(100.0869 g/mol) <em>= 0.1998 ≅ 0.2 mol.</em>
<em></em>
<em>So, the right choice is: B. 0.2.</em>
N=m(g)/m.wt
n=85/12(1)+16(2) =1.93 moles
The molar mass of Li->7g/mol
If 1mol of Li is 7g/mol
1.50mol of Li would be 10.5g/mol
Answer : The mass of of water present in the jar is, 298.79 g
Solution : Given,
Mass of barium nitrate = 27 g
The solubility of barium nitrate at is 9.02 gram per 100 ml of water.
As, 9.02 gram of barium nitrate present in 100 ml of water
So, 27 gram of barium nitrate present in of water
The volume of water is 299.33 ml.
As we know that the density of water at is 0.9982 g/ml
Now we have to calculate the mass of water.
Therefore, the mass of of water present in the jar is, 298.79 g