Consider the follow scenario of 15.3 g of NaCl was dissolved in 155.0 g of water. 1. What is the total mass of the solution?
2. What fraction of the total is Naci?
3. What percent of the total is Naci?
4. Use your percent to determine how many grams of NaCl are contained in 100 g of solution.
5. Determine how many grams of NaCl are in 38.2 g of the solution described at the top of model 1.
6. Use the appropriate two conversion factors to find what volume of this solution you would need to have exactly 2.00g NaCl. The density is 1.07g/mL.

Total mass: 170.3 g

Fraction of NaCl: 0.089%

Percent of NaCl: 8.98%

3.43 g of NaCl in 38.2 g of solution

1 mL . (170.3 g of solution / 1.07 g solution) = 159.1 mL

159.1 mL . (2 g NaCl / 15.3 g NaCl) = 20.8 mL

Explanation:

Our scenario is 15.3 g of NaCl in 155 g of water

Total mass: 15.3 g + 155 g = 170.3 g of solution

Our solute is NaCl - Our solvent is water.

To determine the fraction we divide:

15.3 g / 170.3 g = 0.0898

To determine percent, we multiply the fraction by 100

0.089 . 100 = 8.98 %

We can make a conversion factor to determine the mass of NaCl in 38.2 g of solution. If 15.3 g of NaCl are in 170.3 g of solution and we need 38.2 g, we can propose → (15.3 / 170.3) . 38.2 = 3.43 g of NaCl

The conversion factors are to find what volume of solution is on 2g of NaCl are:

Density data always reffers to solution. So 1.07 grams of solution are contained in 1 mL of solution

1 mL . 170.3 g of solution / 1.07 g solution = 159.1 mL

This is the volume for our 15.3 g of NaCl so:

159.1 mL . (2 g NaCl / 15.3 g NaCl) = 20.8 mL

a saline water is fed to a reverse osmosis unit at flow rate o rotameter connected to reject (retentate) line reads 8 installed in reject and permeate line reads 15.2 milli siemens/cm and 300 micro siemens/com respectively.