Answer : The normality of the solution is, 30.006 N
Explanation :
Normality : It is defined as the number of gram equivalent of solute present in one liter of the solution.
Mathematical expression of normality is:
or,
First we have to calculate the equivalent weight of solute.
Molar mass of solute 
= 94.97 g/mole
Now we have to calculate the normality of solution.
Therefore, the normality of the solution is, 30.006 N
Answer:
982.5 kg/m³
Explanation:
When the temperature of a fluid increases, it dilates, and because of the variation of the volume, it's density will vary too. The density can be calculated by the expression:
ρ₁ = ρ₀/(1 + β*(t₁ - t₀))
Where ρ₁ is the final density, ρ₀ the initial density, β is the constant coefficient of volume expansion, t₁ the final temperature, and t₀ the initial temperature.
At t₀ = 4°C, the water desity is ρ₀ = 1,000 kg/m³. The value of the constant for water is β = 0.0002 m³/m³ °C, so, for t₁ = 93°C
ρ₁ = 1,000/(1 + 0.0002*(93 - 4))
ρ₁ = 1,000/(1+ 0.0178)
ρ₁ = 982.5 kg/m³
(~26grams/mole) and Avogadros # (6.022x10^23) 84.3grams x 1mole/26grams x 6.022x10^23 molecules/mole = 1.95x10^24 molecules of C2H2
The simplified solubility of glucose at 30°C is 1.25 g/g of water. Considering that the density of water at 30°C is 1 g/mL, the equivalent mass of 400 mL of water is also 400g.
The concentration of the solution in water is,
550 g/400g of water = 1.375 g glucose / g of water
Since the concentration is higher compared to the solubility of glucose at the specified temperature, it can be said that the solution is SATURATED.
