Answer:
The answer is 1.15m.
Since molality is defined as moles of solute divided by kg of solvent, we need to calculated the moles of H2SO4 and the mass of the solvent, which I presume is water.
We can find the number of H2SO4 moles by using its molarity
C=nV→nH2SO4=C⋅VH2SO4=6.00molesL⋅48.0⋅10−3L=0.288
Since water has a density of 1.00kgL, the mass of solvent is
m=ρ⋅Vwater=1.00kgL⋅0.250L=0.250 kg
Therefore, molality is
m=nmass.solvent=0.288moles0.250kg=1.15m
Table salt is an ionic compound :)
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Answer:
a) The concentration of drug in the bottle is 9.8 mg/ml
b) 0.15 ml drug solution + 1.85 ml saline.
c) 4.9 × 10⁻⁵ mol/l
Explanation:
Hi there!
a) The concentration of the drug in the bottle is 294 mg/ 30.0 ml = 9.8 mg/ml
b) The drug has to be administrated at a dose of 0.0210 mg/ kg body mass. Then, the total mass of drug that there should be in the injection for a person of 70 kg will be:
0.0210 mg/kg-body mass * 70 kg = 1.47 mg drug.
The volume of solution that contains that mass of drug can be calculated using the value of the concentration calculated in a)
If 9.8 mg of the drug is contained in 1 ml of solution, then 1.47 mg drug will be present in (1.47 mg * 1 ml/ 9.8 mg) 0.15 ml.
To prepare the injection, you should take 0.15 ml of the concentrated drug solution and (2.0 ml - 0.15 ml) 1.85 ml saline
c) In the injection there is a concentration of (1.47 mg / 2.0 ml) 0.735 mg/ml.
Let´s convert it to molarity:
0.735 mg/ml * 1000 ml/l * 0.001 g/mg* 1 mol/ 15000 g = 4.9 × 10⁻⁵ mol/l
Answer: Colligative properties are those properties of solutions that are dependent on the concentration of the solutes in the solution.
Colligative properties has to do with solutions, that is, solutes that are dissolved in solvents. Examples of colligative properties are: freezing point depression, vapour pressure lowering, boiling point elevation and osmotic pressure. Colligative properties do not depend on the identity of the solutes, this implies that the effect of colligative properties are uniform across all solutions. For example, the freezing point depression of any solution will depend on the concentration of solutes that are dissolve in solution.
