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:
19.8% of Nitrogen
Explanation:
In the Al(NO₃)₃ there are:
1 atom of Al
3 atoms of N
And 9 atoms of O
The molar mass of Al(NO₃)₃ is:
1 Al * (26.98g/mol) = 26.98g/mol
3 N * (14g/mol) = 42g/mol
9 O * (16g/mol) = 144g/mol
26.98 + 42 + 144 = 212.98g/mol
We can do a conversion using these molar masses to find the mass of nitrogen is the sample, that is:
2.57g * (42g/mol / 212.98g/mol) =
0.51g N
Percent composition of nitrogen is:
0.51g N / 2.57g * 100
= 19.8% of Nitrogen
Answer:
sample A
Explanation:
the first one because of the ppm value