The items are answered below and are numbered separately for each compound.
The freezing point of impure solution is calculated through the equation,
Tf = Tfw - (Kf)(m)
where Tf is the freezing point, Tfw is the freezing point of water, Kf is the freezing point constant and m is the molality. For water, Kf is equal to 1.86°C/m. In this regard, it is assumed that m as the unit of 0.25 is molarity.
1. NH4NO3
Tf = 0°C - (1.86°C/m)(0.25 M)(2) = -0.93°C
2. NiCl3
Tf = 0°C - (1.86°C/m)(0.25 M)(4) = -1.86°C
3. Al2(SO4)3
Tf = 0°C - (1.86 °C/m)(0.25 M)(5) = -2.325°C
For boiling points,
Tb = Tbw + (Kb)(m)
For water, Kb is equal to 0.51°C/m.
1. NH4NO3
Tb = 100°C + (0.51°C/m)(0.25 M)(2) = 100.255°C
2. NiCl3
Tb = 100°C + (0.51°C/m)(0.25 M)(4) = 100.51°C
3. Al2(SO4)3
Tb = 100°C + (0.51°C/m)(0.25 M)(5) = 100.6375°C
Answer:
0.971 grams
Explanation:
Given:
Temperature = 3.0° C = 3 + 273 = 276 K
Volume, V = 5.0 L
Pressure, P = 0.100 atm
Now, from the relation
PV = nRT
where,
n is the number of moles,
R is the ideal gas constant = 0.082057 L atm/mol.K
thus,
0.1 × 5 = n × 0.082057 × 276
or
n = 0.022 moles
Also,
Molar mass of the Dinitrogen monoxide gas (N₂O)
= 2 × Molar mass of nitrogen + 1 × Molar mass of oxygen
= 2 × 14 + 16 = 44 grams/mol
Therefore, Mass of 0.022 moles of N₂O = 0.022 × 44 = 0.971 grams
the transition of a substance directly from the solid to the gas phase, without passing through the intermediate liquid phase.
Answer:
0.077 M
Explanation:
Molarity is the representation of the solution.
Molarity:
It is amount of solute in moles per liter of solution and represented by M
Formula used for Molarity
M = moles of solute / Liter of solution . . . . . . . . . . (1)
Data Given :
The concentration of half normal (NaCl) saline = 0.45g / 100 g
So,
Volume of Solution = 100 g = 100 mL
Volume of Solution in L = 100 mL / 1000
Volume of Solution = 0.1 L
molar mass of NaCl = 58.44 g/mol
Now to find number of moles of Nacl
no. of moles of NaCl = mass of NaCl / molar mass
no. of moles of NaCl = 0.45g / 58.44 g/mol
no. of moles of NaCl = 0.0077 g
Put values in the eq (1)
M = moles of solute / Liter of solution . . . . . . . . . . (1)
M = 0.0077 g / 0.1 L
M = 0.077 M
So the molarity of half-normal saline solution (0.45% NaCl) = 0.077 M
