The volume of 0.150 M NaHCO_3 (FW = 84.0) solution that contains 6.30 g of NaHCO

Question

The volume of 0.150 M NaHCO_3 (FW = 84.0) solution that contains 6.30 g of NaHCO_3 is a) 0.200 L. b) 0.400 L. c) 0.500 L. d) 1.60 L. e) 2.50 L. How many mL 0.540 M NaCl (saline) can be prepared by diluting 100 mL of a 6.00 M NaCl solution? a) 196 b)98.0 c)900. d) 24.0 e) 1110 Given a specific volume of a solution of 0.5555 M HClO_4, the best way to determine the number of moles is to. a) multiply the vol. by the molecular mass and divide by the concentration b) multiply the concentration by the molecular mass and divide by the volume c) multiply the concentration by the volume d) divide the concentration by the molecular mass and volume e) divide by the volume and multiply by the molecular mass f) None of the above.

Explanation / Answer

1) molar mass of NaHCO3 = 84.0 g/mol

find first mole of NaHCO3 = (6.30 g/ 84.0 g/mol) = 0.075 moles

molarity = number of moles / volume in Litter

0.150 M = 0.075 moles / V

V = 0.500 L

2) find number of moles of NaCl in 100.0 mL of a 6.00 M

moles = 6.00 M x 0.1 L = 0.6 moles

find volume of diluting solution of molarity = 0.540 M

volume = 0.6 moles / 0.540 M = 1.11 L = 1110 mL

3) option c is correct

if we have given molarity and volume of solution than simply use

Molarity = no. of moles / volume in Litter

molarity is concentration of solution than simply

concentration multiply by volume get moles

Concentration (molarity) x volume = moles

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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