Question

need help with questions 3,4,5

Laboratory Questions: Saturated Solutions The data below was collected in an experiment to determine the solublity of sodium nitrate at 20 C. Total Volume of Water (mL) Saturation Temp (C Solubility (g salt/100g water) 3.50 139.3 78.0 3.75 70.0 4.00 60.0 121.8 4.25 52.0 4.50 45.0 08. 3 4.75 5.00 36.0 Na NO3 73 1. If 4.8739 g of NaNO, was used, calculate the solubility in units of g NaNO,/100 g water at each saturation temperature. Show your first calculation here. Complete the rest of the calculations and fill in the table oog of 20 134.2542 glgz On the attached piece of graph paper, construct a graph of solubility as a function of saturation temperature 2.

Answer 1

Ans 3 :

The solubilty of sodium nitrate 20^oC from the graph is : 85 g / 100 g water

Ans 4 :

85 g of salt dissolves in 100 g water at 20 degrees celsius.

Let the mass of water be 100 g , so mass of salt will be 85 g

mass of solution = 100 g + 85 g = 185 g\

mass percent = ( mass salt / mass solution ) x 100

= ( 85 g / 185 g ) x 100

= 45.9 %

Ans 5 :

density = mass / volume

1.32 g/mL = 185 g / V

V = 140.15 mL

Mol NaNO3 = mass /molar mass

= 85 g / 84.9947 g/mol = 1 mol

Molarity = mol / volume of solution (L)

= 1 mol / 0.14015 L

= 7.13 M