Answer:
Molality = 0.0862 mole/kg
Explanation:
Molality = (number of moles of solute)/(mass of solvent in kg)
Number of moles of solute = (mass of Creatinine in the blood sample)/(Molar mass of Creatinine)
To obtain the mass of creatinine in 10 mL of blood. We're told that 1 mg of Creatinine is contained in 1 decilitre of blood.
1 decilitre = 100 mL
1 mg of Creatinine is contained in 100 mL of blood
x mg of Creatinine is contained in 10 mL of blood.
x = (1×10/100) = 0.1 mg = 0.0001 g
Molar mass of Creatinine (C₄H₇N₃O) = 113.12 g/mol
Number of moles of Creatinine in the 10 mL blood sample = (0.0001/113.12) = 0.000000884 moles
Mass of 10 mL of blood = density × volume = 1.025 × 10 = 10.25 mg = 0.01025 g = 0.00001025 kg
Molality of normal creatinine level in a 10.0-ml blood sample = (0.000000884/0.00001025)
Molality = 0.0862 moles of Creatinine per kg of blood.
Hope this Helps!!!
3) +7
I got this stupid question wrong so that you people don't have to. I simply hate chemistry and I wish it didn't exist. Kind of like this website that makes me explain the answer.
Trust me, the answer is correct
<span>Who can help me with a few chemistry balancing equations?
Me. Just show me the equations :)</span>
