There are 2071.79 grams of glucose in 11.5 moles.
Answer:
2.1 × 10⁻¹ M
2.0 × 10⁻¹ m
Explanation:
Molarity
The molar mass of aniline (solute) is 93.13 g/mol. The moles corresponding to 3.9 g are:
3.9 g × (1 mol/93.13 g) = 0.042 mol
The volume of the solution is 200 mL (0.200 L). The molarity of aniline is:
M = 0.042 mol/0.200 L = 0.21 M = 2.1 × 10⁻¹ M
Molality
The moles of solute are 0.042 mol.
The density of the solvent is 1.05 g/mL. The mass corresponding to 200 mL is:
200 mL × 1.05 g/mL = 210 g = 0.210 kg
The molality of aniline is:
m = 0.042 mol/0.210 kg = 0.20 m = 2.0 × 10⁻¹ m
C.) hydrogen bonding interactions.
Answer : The mass of lithium hypochlorite are, 34.7 grams.
Explanation : Given,
Moles of
= 0.594 g
Molar mass of
= 58.4 g/mol
Expression used :
Now put all the given values in this expression, we get:
Therefore, the mass of lithium hypochlorite are, 34.7 grams.