Answer:
Explanation:
molecular weight of creatinine = 131
1 mg of creatinine = 1 x 10⁻³ / 131 = 7.63 x 10⁻⁶ mole of creatinine.
volume of solution = .1 L
mass of blood solution = .1 x 1025
= 102.5 g
mass of solvent = 102.5 g approximately
= .1025 kg
molality = mole of solute / mass of solvent in kg
= 7.63 x 10⁻⁶ / .1025 kg
= 74.44 x 10⁻⁶ .
Osmotic pressure :---
π V / T = nR π is osmotic pressure , V is volume of solution in liter , T is absolute temperature , n is molality .
π x .1 / 298 = 74.44 x 10⁻⁶ x .082
π = 18.19 x 10⁻³ atm
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>
Answer:
1 957,5
Explanation:
87libs=39.5kg that equals 50=1.957.5
