Answer:
- 0.0413°C ≅ - 0.041°C (nearest thousands).
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>
Because it has high electrical conductivity
Can you show me a picture or a diagram
23.67 g unless you're rounding to significant figures than it's 24 g
Mass = Density x Volume
Mass = 3.2g/mL x 5 mL
Mass = 16g
