When 136g of glycine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezing point of pure . On the other hand, when of sodium chloride are dissolved in the same mass of , the freezing point of the solution is lower than the freezing point of pure . Calculate the van't Hoff factor for sodium chloride in . Be sure your answer has a unit symbol, if necessary, and round your answer to significant digits.

The given question is incomplete. The complete question is:

When 136 g of glycine are dissolved in 950 g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 136 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for sodium chloride in X.

The vant hoff factor for sodium chloride in X is 1.9

Explanation:

Depression in freezing point is given by:

= Depression in freezing point

= freezing point constant

i = vant hoff factor = 1 ( for non electrolyte)

m= molality =

Now Depression in freezing point for sodium chloride is given by:

= Depression in freezing point

= freezing point constant  

m= molality =

Thus vant hoff factor for sodium chloride in X is 1.9

here is table 2-2:

country

percent of eligible voters that voted

rank

united states

44.9%

138

canada

60.1%

107

costa rica

84.6%

14

hungary

66.9%

82

japan

57.0%

118

egypt

27.7%

159

south africa

85.5%

10

ans is no, the data set cannot be plotted as a single circle graph

weight of any object is given by

w = m*g

where m is mass of the object and g is the acceleration due to gravity.

for earth the value of g is 9.8 m/s^2 so

w=117.6 n m=12 g=?

117.6=12g => g=9.8 m/s^2

which is the force of gravity on earth.