An aqueous solution of magnesium sulfate is prepared by dissolving 3.81 g of mag

Question

An aqueous solution of magnesium sulfate is prepared by dissolving 3.81 g of magnesium sulfate in 7.32 times 102 g of water. The density of the solution is 1.04 g mL-1. Determine the mass percent of magnesium sulfate in the solution. Determine the mole fraction of magnesium sulfate in the solution. An aqueous solution of sodium chlorite is prepared by dissolving 7.50 g of sodium chlorite in 3.29 times 102 g of water. The density of the solution is 1.49 g mL -1. Determine the molarity (in mol/L) of sodium chlorite in the solution.

Explanation / Answer

total mass of solution = 3.81 + 732 = 735.81

a) mass percent of mgso4 = 3.81 / 735.81 * 100

= 0.5178 %

b) molar mass of mgso4 = 24 + 32 + 64 = 120 g

so, no of moles of MgSoO4 = 3.81 / 120 = 0.03175

molar mass of H2O = 18 g

no. of moles of water = 732/18 = 40.667

total no. of moles = 40.667 + 0.03175

= 40.6984177

mole fractuon of MgSO4 = 0.03175 / 40.6984167 * 100

= 0.078013 %

2)

In 1 litre of the solution ,

mass of solution = 1.49 * 1000 = 1490 g

7.5 g is dissolved in 329 gram water

so, total mas of solution = 329 + 7.5 = 336.5

so, in 336.5 gm of solution , water = 7.5 g

In 1490 gram of sol, water = 7.5 * 1490 /336.5 = 33.20951

molar mass of NaCl = 58.5 grams

so, no. of moles of NaCl = 33.20951 / 58.5 = 0.567684

so, molarity = 0.567684 Moles / liter

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.