Question

MULTIL HULI U vull a la volume UI U. JUUL. PQ-4. What solution volume is needed to prepare a 0.110 M solution of MgSO. using 10.3 g of the compound? The molar mass of MgSO, is 120.366 g-moll. (A) 9.41*10'L (B) 0.778 L (C) 1.11 L (D) 93.6L PO-5. What is the molar concentration of a solution made by dissolving 8.56 g of sodium acetate in water and diluting to 750.0 mL? The molar mass of NaC,H,O, is 82.03 g mol"!. (A) 5.30 M (B) 0.139 M (C) 0.104 M (D) 0.0783 M PO-6. What is the molar concentration of the chloride ion in a 3.0 M CaCl solution? (A) 6.0 M (B) 3.0 M (C) 1.5 M (D) 1.0M Conceptual PO-7. What is the concentration of the bromide ion if 25.0 mL of a 0.50 M AB solutie 40.0 mL of a 0.35 M NaBr solution? Assume volumes are additive. (A) 7.9x10molL- (B) 0.41 mol L (C) 0.79 mol-L- (D) 2.1 mol L- What is the final concentration when 28.2 mL of an 11.8 M solution is diluted to 1000 m 2 (A) 0.333 M (B) 2.82 M (C) 3.33 M (D) 4.18 M

6,7,8

Answer 1

Question 6

I mole of CaCl2 contains 2 moles of Chloride ions.

Therefore, if the solution is 3.00M CaCl2, then there will be twice as many chloride ions, i.e. 6.00 M.

Question 7.

Number of moles of bromide ion in AlBr3 solution:

For 1 mole AlBr3, there will be 3 moles of Bromide ions

Number of moles of AlBr3 = concentration × volume

= 0.50M x (25/1000)L

= 0.0125 moles of AlBr3

Therefore number of moles of Bromide ions= 3 x 0.0125 = 0.0375 moles.

Number of moles of NaBr = concentration x volume

= 0.35M × (40/1000)L (Dividing by 1000 to convert mL to L)

= 0.014 moles of NaBr

I mole of NaBr contains 1 bromide ion. Therefore,c0.014 moles of NaBr contain 0.014 moles of bromide ion.

So, after mixing both solutions total number of moles of bromide ion = 0.0375 + 0.014 = 0.0515 moles

Concentration = number of moles / total volume.

= 0.0515/((25+40)/1000) = 0.79 mol/L

Therefore answer is (c).

Question 8.

Number of moles = concentration x volume

= 11.8 x (28.2/1000)

= 0.33276

After diluting to 100mL, the final volume changes but the number of moles remains constant. So the new concentration is

Concentration = number of moles/volume

= 0.33276 /(100/1000)

= 3.33 M (option c)