Ascientist wants to make a solution of tribasic sodium phosphate, na3po4, for a laboratory experiment. how many grams of na3po4 will be needed to produce 725 ml of a solution that has a concentration of na+ ions of 1.00 m ? express your answer to three significant figures and include the appropriate units.

Answer :

The correct answer is for mass of Na₃PO₄ 39.7 g.

Given  : 1) Molarity of Na⁺ ions = 1.00 M or 1.00

2) Volume of solution = 725 mL

Converting volume of solution from mL to L :

Conversion factor : 1 L = 1000 mL  

Volume of solution = 0.725 L

Following steps can be  done to find mass of  :

Step 1 : Write the dissociation reaction of Na₃PO₄ .

Step 2:  Find moles of Na⁺ ions :

Mole of Na⁺ ions can be calculated using molarity formula  which is :

Plugging value of Molarity and volume

Multiplying both side by 0.725 L

Mole of Na⁺ ions = 0.725 mol

Step 3: Find mole ratio of Na₃PO₄ : Na⁺ :

Mole ratio  is found from coefficients from balanced reaction as:

Mole of Na₃PO₄ in balanced reaction = 1

Mole of Na⁺ ion =  3

Hence mole ratio of Na₃PO₄: Na⁺ = 1 : 3

Step 4 : To find mole of Na₃PO₄

Mole of Na₃PO₄ can be calculated using mole of Na⁺ ion and Mole ratio as :

Mole of Na₃PO₄ =  0.242 mol

Step 5 : To find mass of Na₃PO₄

Mole of Na₃PO₄ can be converted to mass of Na₃PO₄ using  molar mass of Na₃PO₄ as :

Mass of Na₃PO₄ = 39.619 g

Step 6 : To round off mass of Na₃PO₄ to correct sig fig .

The sig fig in 750 mL and 1.00 M is 3 . So mass of Na₃PO₄ can rounded to 3 sig fig as :

Mass of Na₃PO₄ = 39.7 g

8

explanation:

each principal energy level above the first contains one s orbital and three p orbitals. a set of three p orbitals, called the p sublevel, can hold a maximum of six electrons. therefore, the second level can contain a maximum of eight electrons - that is, two in the s orbital and 6 in the three p orbitals.

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