THE ANSWER TO THE QUESTION IS A
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The question incomplete , the complete question is:
A student dissolves of 18.0 g urea in 200.0 mL of a solvent with a density of 0.95 g/mL . The student notices that the volume of the solvent does not change when the urea dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to significant digits.
Answer:
The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.
Explanation:
Moles of urea = 
Volume of the solution = 200.0 mL = 0.2 L (1 mL = 0.001 L)

Molarity of the urea solution ;

Mass of solvent = m
Volume of solvent = V = 200.0 mL
Density of the urea = d = 0.95 g/mL


(1 g = 0.001 kg)
Molality of the urea solution ;


The molarity and molality of the student's solution is 1.50 Molar and 1.58 molal.
Answer:
None of the given options
Explanation:
Let's go case by case:
A. No matter the volume, the concentration of Fe(NO₃)₃ (and thus of [Fe³⁺] as well) is 0.050 M.
B. We can calculate the moles of Fe₂(SO₄)₃:
- 0.020 M * 0.80 L = 0.016 mol Fe₂(SO₄)₃
Given that there are two Fe⁺³ moles per Fe₂(SO₄)₃ mol, in the solution we have 0.032 moles of Fe⁺³. With that information in mind we <u>can calculate [Fe⁺³]</u>:
- 0.032 mol Fe⁺³ / 0.80 L = 0.040 M
C. Analog to case A., the molar concentration of Fe⁺³ is 0.040 M.
D. Similar to cases A and C., [Fe⁺³] = 0.010 M.
Thus none of the given options would have [Fe⁺³] = 0.020 M.