Answer : The correct answer for change in freezing point = 1.69 ° C
Freezing point depression :
It is defined as depression in freezing point of solvent when volatile or non volatile solute is added .
SO when any solute is added freezing point of solution is less than freezing point of pure solvent . This depression in freezing point is directly proportional to molal concentration of solute .
It can be expressed as :
ΔTf = Freezing point of pure solvent - freezing point of solution = i* kf * m
Where : ΔTf = change in freezing point (°C)
i = Von't Hoff factor
kf =molal freezing point depression constant of solvent.
m = molality of solute (m or 
)
Given : kf = 1.86 
m = 0.907 )
Von't Hoff factor for non volatile solute is always = 1 .Since the sugar is non volatile solute , so i = 1
Plugging value in expression :
ΔTf = 1* 1.86 * 0.907 )
ΔTf = 1.69 ° C
Hence change in freezing point = 1.69 °C
Explanation and answer:
The molar volume of a gas is 22.4 L at 0 deg. C.
So the molar mass of the gas is, by proportion,
0.16 g * (22400 mL)/(240 mL) = 14.93 g
The molar mass of (CH4)n = 12+4(1) = 16.
So n = 15.93/16 = 1, or the molecular formula is CH4.
Note: The temperature at which the volume was observed was not given. If 240 cm^3 was observed at 20 deg.C, then the volume at 0 deg.C would be
V=240*(273+0)/(273+20) = 223.6
The molar mass = 0.16*22400/(223.6) = 16.03
which gives n = 16/16.03 = 1 again, but more accurately.
Answer:
CH2O
Explanation:
Firstly, we need to convert the masses of the elements to percentage compositions. This can be done by placing the mass of each element over the total mass multiplied by 100% . We can start with carbon.
C = 5.692/14.229 * 100 = 40%
O = 7.582/14.229 * 100 = 53.29%
H = 0.955/14.229 * 100 = 6.71%
We then proceed to divide each percentage composition by their atomic mass of 12, 16 and 1 respectively.
C = 40/12 = 3.333
O = 53.29/16 = 3.33
H = 6.71/2 = 6.71
Dividing by the smaller value which is 3.33
C = 3.33/3.33 = 1
O = 3.33/3.33= 1
H = 6.71/3.33 = 2
The empirical formula of the compound ribose is CH2O
0.11031 mg
I think that is the answer
