Answer to Question #345833 in General Chemistry for Jonalyn Galao

Question

3.)calculate the molality of phosphoric acid, H3PO4, in a solution of 29.0g H3PO4 in 250.0g H2O. (atomic mass :H=1.008g/mol:P=30.97g/mol:O=16.00g/mol:Ans.m=1.18mol solute /kg solvent)

4.)Refer to the data given in problem #3, calculate the mole fraction of H3PO4 and H20. (Ans. X H3PO4 =0.979:XH20=0.0209

5.)An aqueous solution has 0.0400g of NaCl (salute) in 2.00kg H2O(solvent). What is the concentration of the solution in parts per million

6.)calculate the volume, in mL, of 1.72M HCl solution that will react with 2.67 mol of CaCO3? (Ans. 3.10x103mL)

__HCl+__CaCl2+__H2O+__CO2

7.)If 0.650% solution of Na2CO3 will be used to precipitate Ca+2 from solution, determine the amount of solution, in grams, needed to precipitate 5.00x102mL of 0.01120M Ca+? (Atomic mass :Na 22.99g/mol:C=12.01g/mol:O=16.00g/mol:Ca=40.08g/mol:Ans.91.3g

___Na2CO3(aq)+__Ca+2(aq)___CaCO3(s)+___Na+(aq)

Answer 1

3.)

Concentration of molality = solute(mol) / solvent(kg)

n (H₃PO₄) = m / M_r = 29 / 98 = 0,296 mol

n — amount of substance (mol);

m — mass of substance (g);

M_r — relative molecular mass (g/mol)

Molality= 0,296 mol / 0,25 kg =1,184 mol/kg

ANSWER: 1,184 mol/kg

4.)

n (H₃PO₄) = 0,296 mol

n (H₂O) = m / M_r = 250 / 18 = 13,89 mol

n (total) = 0,296 + 13,89 = 14,185 mol

X (H₃PO₄) =0,296 mol / 14,185 mol = 0,0209

X (H₂O) = 13,89 mol / 14,185 mol = 0,9791

ANSWER:

H₂O : 0,09791

H₃PO₄ : 0,0209

5.)

the density (p) of water is 1 g/ml.

m (NaCl) = 0,04 g = 40 mg

V (H₂O) = m * p = 2kg * 1 = 2 litr

ppm = m (mg) / V (l) = 40 / 2 = 20

ANSWER: ppm= 20

6).

CaCO3 + 2 HCl = CaCl₂ + CO₂ + H2O

According to the above reaction,

1 mol CaCO₃ —> 2 mol HCl

2,76mol CaCO₃ —> X

X = 2,76 * 2 / 1 = 5,52 mol HCl

V = n / C_M = 5,52mol / 1,72mol/l = 3,21 litr or 3,21 * 10³ ml

V — volume of solution (litr);

n — amount of substance (mol);

C_M — concentration of molarity.

ANSWER: 3,21 * 10³ ml.

7).

a)

V — volume = 5*10²ml = 0,5 litr;

C_M — concentration of molarity =

= 0,0112 mol/l;

n — amount of substance (mol) = ?

n (Ca²⁺) = V * C_M = 0,5 * 0,0112 =0,0056 mol

b)

M_r (Na₂CO₃) = 23*2+12+16*3 = 106 g/mol

Na₂CO₃ + Ca²⁺ => CaCO₃ + 2Na⁺

According to the above reaction,

106 g Na₂CO₃ — > 1mol Ca²⁺

X — > 0,0056mol Ca²⁺

X = 106 * 0,0056 / 1 = 0,5936 g Na₂CO₃

c)

0,5936 g — 0,65%

X — 100%

X = 0,5936 * 100 / 0,65 = 91,3 g

ANSWER: 91,3 g