molar concentration of AgNO₃ solution = 0.118 mole/L
Explanation:
Because we have the volume of the solution and there is no information about the density of the solution I will asume that you ask for the molar concentration.
number of moles = mass / molecular weight
number of moles of AgNO₃ = 10 / 170 = 0.0588
molar concentration = number of moles / volume (L)
molar concentration of AgNO₃ solution = 0.0588 / 0.5
molar concentration of AgNO₃ solution = 0.118 mole/L
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<u>Answer:</u>
For 2: The correct answer is grams of solute per 100 grams of solvent.
For 3: The correct answer is supersaturated.
For 4: the correct answer is the solubility decreases.
<u>Explanation:</u>
Solubility is defined as the property which refers to the ability of the solute that can be dissolved in a solvent. It is defined as the number of grams of solute per 100 grams of solvent.
Unsaturated solution is defined as the solution in which amount of solute that is dissolved in the solvent is less.
Saturated solution is defined as the solution in which no more solute can be dissolved in the given amount of solvent.
Emulsion is defined as the dispersion of one liquid in another liquid in which it is not soluble.
Supersaturated solution is defined as the solution in which solvent contains more amount of solute than the required amount. These solutions help in the process of crystallization.
When a crystal is added to a <u>supersaturated solution</u>, more and more particles come out of the solution and this process is known as crystallization.
According to the Henry's Law
The solubility of the gas in a liquid is directly proportional to the partial pressure of the gas.

With increase in the partial pressure, the solubility of the gas in liquid also increases and vice-versa.
Hence, the correct answer is the solubility decreases.
Answer:
<em>- 0.0413°C ≅ - 0.041°C (nearest thousands).</em>
Explanation:
- Adding solute to water causes the depression of the freezing point.
<em>ΔTf = Kf.m,</em>
Where,
ΔTf is the change in the freezing point.
Kf is the freezing point depression constant (Kf = 1.86 °C/m).
m is the molality of the solution.
<em>Molality is the no. of moles of solute per kg of the solution.</em>
- <em>no. of moles of solute (glucose) = mass/molar mass</em> = (8.44 g)/(180.156 g/mol) = <em>0.04685 mol.</em>
<em>∴ molality (m) = no. of moles of solute/kg of solvent</em> = (0.04685 mol)/(2.11 kg) = <em>0.0222 m.</em>
∴ ΔTf = Kf.m = (1.86 °C/m)(0.0222 m) = 0.0413°C.
<em>∴ The freezing point of the solution = the freezing point of water - ΔTf </em>= 0.0°C - 0.0413°C = <em>- 0.0413°C ≅ - 0.041°C (nearest thousands).</em>
N=m(g)/m.wt
n=85/12(1)+16(2) =1.93 moles