Density of the mixture = mass of the mixture / volume of the mixture
Mass of the mixture = mass of antifreeze solution + mass of water.
Mass of antifreeze solution = density of the antifreeze solution * volume
Mass of antifreeze solution = 0.8g/ml * 5.1 l * 1000 ml / l = 4,080 g
Mass of water = density of water * volume of water = 1.0 g/ml * 3.8 l * 1000 ml / l = 3,800 g
Mass of mixture = 4080 g + 3800 g= 7880 g
Volume of mixture = volume of antifreeze solution + volume of water
Volume of mixture = 5100 ml + 3800 ml = 8900 ml
Density of mixture = 7800 g / 8900 ml = 0.876 g/ml
Specific gravity of the mixture = density of the mixture / density of water = 0.876
Answer: 0.876
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Explanation:
Given :
Amount of solute - sucrose (C12H22O11) = 41 g
Amount of solvent -soda = 355-mL
Molarity of the solution with respect to sucrose= ?
Molarity(M) is a unit of concentration measuring the number of moles of a solute per liter of solution. The SI unit of molarity is mol/L.
Formula to find the molarity of solution :
Molarity =
Amount of solvent is given in mL, let’s convert to L :
1 L = 1000 mL
Therefore, 355 mL in L will be :
= 0.355 L
We have the amount of solute in g, let’s calculate the number of moles first :
Number of moles (n) =
Molar mass of C12H22O11 = 342.29 g/mol.
Therefore, n =
= 0.119 moles.
Answer:
The maximum pressure is 612.2 Pa
Explanation:
The pressure of the ice (P1) = 624 Pa
The temperature of the ice = 273.16 K
The maximum temperature the specimen = - 5 oC
= -5 + 273 = 268K
The maximum Pressure the freeze drying can be will be (P2) = ?
Using Pressure law, which shows the relationship between pressure and temperature.
P1 / T1 = P2 / T2
P2 T1 = P1 T2
P2 = P1 T2 / T1
P2 = 624 × 268 / 273.16
P2 = 612.2 Pa
The maximum pressure at which drying can be carried out is 612.2 Pa
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