Acertain liquid x has a normal freezing point of 7.60 °c and a freezing point depression constant k= 6.90 °c-kg-mol. calculate the freezing point of a solution made of 7.57g of sodium chloride (nacl) dissolved in 350. g of x round your answer to 3 significant digits. lºc x 5

The freezing point of solution is -5.11°C

Explanation:

Vant hoff factor for ionic solute is the number of ions that are present in a solution. The equation for the ionization of sodium chloride follows:

The total number of ions present in the solution are 2.

To calculate the molality of solution, we use the equation:

Where,

= Given mass of solute (NaCl) = 7.57 g

= Molar mass of solute (NaCl) = 58.44 g/mol

= Mass of solvent (liquid X) = 350.0 g

Putting values in above equation, we get:

To calculate the depression in freezing point, we use the equation:

where,

i = Vant hoff factor = 2

= molal freezing point depression constant = 6.90°C/m

m = molality of solution = 0.370 m

Putting values in above equation, we get:

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

= 5.11 °C

Freezing point of water = 0°C

Freezing point of solution = ?

Putting values in above equation, we get:

Hence, the freezing point of solution is -5.11°C

h let me think about that

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.