A saturated solution contains more solute than solvent. contains more solvent th

Question

A saturated solution contains more solute than solvent. contains more solvent than solute. contains equal moles of solute and solvent. contains the maximum amount of solute that will dissolve in that solvent at that temperature. Contains a solvent with only sigma bonds and no pi bonds (i.e. only single bonds, with no double or triple bonds). Which response lists all the following pairs that are miscible liquids? octane (Cillis) and water acetic acid (CH'COOH) and water octane (Callia) and carbon tetrachloride (CCU) A 9.50 % by mass solution of acetone (C_2H_6O) in water has a density of 0.9849 g ml. at 20degreeC What is the molarity of this solution? What is the molality of a solution that is 3.68 % by mass calcium chloride? Calculate the molality of 6.0 M H:SO i solution. The density of the solution is 1.34 g/mL. The solubility of nitrogen gas at 25"C and a nitrogen pressure of 522 mmHg is 4.7 x 10^-4 moll.. What is the value of the Henry's Law constant in mo LI. aim? According to Raoult's law. which statement is false! The vapor pressure of a solvent over a solution decreases as its mole fraction increases. The solubility of a gas increases as the temperature decreases. The vapor pressure of a solvent over a solution is less than that of pure solvent. The greater the pressure of a gas over a solution the greater its solubility. Ionic solutes dissociate in solution causing an enhancement of all colligating properties. Calculate the freezing point of a solution made from 22.0 g of octane (C_8H_18) dissolved in 148.0 g of benzene. Benzene freezes at 5.50*C and its Ki value is 5.12degreeC/m.

Explanation / Answer

1) d

2) e

3) B

Taking 100 gm of solution, volume of solution = 100/density (v) = 101.53 ml

mass of acetone = 9.5 gm

moles of acetone = 9.5/ 58 = 0.164

so molarity of solution = 0.164/v = 1.61 M

4) B

Let mass of solution be 100 gm = 0.1 kg.

mass of CaCl2 = 3.68 gm

moles of CaCl2 = 3.68/111 = 0.0332

molality = 0.0332/0.1 = 0.332

5) A

Volume of 100 gm or 0.1 kg solution = 74.63 ml

moles of H2SO4 taken = 6 x 74.63 = 447.78 mmoles

molality = 447.78/100 = 4.48 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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